Understanding Elastic and Inelastic Collisions - AP Physics C: Electricity and Magnetism

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Question

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

$v_$1^{}$=final\ velocity\ of\ ball\ 1$

$v_$2^{}$=final\ velocity\ of\ ball\ 2$

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Answer

We must use conservation of momentum to tackle this problem.

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ $\text{kg}$$

$v_1=7.1\ $\frac{m}{s}$$

and we know the following for the second ball:

$m_2=0.95\ $\text{kg}$$

$v_2=-2.8\ $\frac{m}{s}$$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

Plug this into our conservation of momentum equation.

So now is eliminated and we can solve for .

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$

$v_2'=\frac{2(0.54\ \text{kg}$)(7.1\ $\frac{m}{s}$)+(0.95\ $\text{kg}$-0.54\ $\text{kg}$)(-2.8\ $\frac{m}{s}$)}{0.54\ $\text{kg}$+0.95\ $\text{kg}$}$

$v_2'=4.4\ $\frac{m}{s}$$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

$v_1'=4.4\ $\frac{m}{s}$-7.1\ $\frac{m}{s}$-2.8\ $\frac{m}{s}$$

$v_1'=-5.5\ $\frac{m}{s}$$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Question

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

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Answer

For the 30 kg cart, we know

$m_1=30\ $\text{kg}$$

$v_1=9\ $\frac{m}{s}$$

and for the 46 g cart, we know

$m_2=46\ $\text{kg}$$

$v_2=-4\ $\frac{m}{s}$$ .

After the collision, we have .

Use conservation of momentum to solve this problem.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

$v=\frac{(30\ \text{kg}$)(9\ $\frac{m}{s}$)+(46\ $\text{kg}$)(-4\ $\frac{m}{s}$)}{30\ $\text{kg}$+46\ $\text{kg}$}$

$v=1.13\ $\frac{m}{s}$$

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Question

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

Tap to reveal answer

Answer

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ $\text{kg}$$ ,

Female skater: $m_2=60\ $\text{kg}$$ ,

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_$2v_{2x}$=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_$1v_{1y}$=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{\sin\theta}{\cos\theta}$

$$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\tan\theta$

Plug in the numbers to find the angle

$$\frac{(68\ \text{kg}$)(15\ $\frac{m}{s}$)}{(60\ $\text{kg}$)(12\ $\frac{m}{s}$)}=\tan\theta$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_$2v_{2x}$=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_$2v_{2x}$$}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg}$)(12\ $\frac{m}{s}$)}{(68\ $\text{kg}$+60\ $\text{kg}$)\cos(54.8)}$

$v=9.8\ $\frac{m}{s}$$

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Question

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0$\frac{m}{s}$$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0$\frac{m}{s}$$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

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Answer

Relevant equations:

$K = $$\frac{1}{2}$mv^2$$

$\Delta K = K_f - K_i$

Use conservation of momentum to determine the final velocity of the objects.

$m_$Av_{A,i}$+m_$Bv_{B,i}$=m_$Av_{A,f}$+m_$Bv_{B,f}$$

$m_$Av_{A,i}$+m_$Bv_{B,i}$=(m_A + m_B)v_f$

$(1kg)(6$\frac{m}{s}$)+(5kg)(4$\frac{m}{s}$)=(1kg+5kg)v_f$

$v_f=4.33$\frac{m}{s}$$

Calculate the total initial kinetic energy of the two objects.

$K_${B,i}=\frac{1}{2}$(5kg)(4$\frac{m}{s}$)^2$=40J$

$K_i = $K_{A,i}$$+K_{B,i}$=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_${A,f}=\frac{1}{2}$(1kg)(4.33$\frac{m}{s}$)^2$=9.39J$

$K_${B,f}=\frac{1}{2}$(5kg)(4.33$\frac{m}{s}$)^2$=46.94J$

$K_f = $K_{A,f}$$+K_{B,f}$=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$$\frac{|\Delta K|}{K_i}$100% = $\frac{1.67}{58}$100% \approx 2.9%$

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Question

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

$v_$1^{}$=final\ velocity\ of\ ball\ 1$

$v_$2^{}$=final\ velocity\ of\ ball\ 2$

Tap to reveal answer

Answer

We must use conservation of momentum to tackle this problem.

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ $\text{kg}$$

$v_1=7.1\ $\frac{m}{s}$$

and we know the following for the second ball:

$m_2=0.95\ $\text{kg}$$

$v_2=-2.8\ $\frac{m}{s}$$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

Plug this into our conservation of momentum equation.

So now is eliminated and we can solve for .

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$

$v_2'=\frac{2(0.54\ \text{kg}$)(7.1\ $\frac{m}{s}$)+(0.95\ $\text{kg}$-0.54\ $\text{kg}$)(-2.8\ $\frac{m}{s}$)}{0.54\ $\text{kg}$+0.95\ $\text{kg}$}$

$v_2'=4.4\ $\frac{m}{s}$$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

$v_1'=4.4\ $\frac{m}{s}$-7.1\ $\frac{m}{s}$-2.8\ $\frac{m}{s}$$

$v_1'=-5.5\ $\frac{m}{s}$$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Question

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

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Answer

For the 30 kg cart, we know

$m_1=30\ $\text{kg}$$

$v_1=9\ $\frac{m}{s}$$

and for the 46 g cart, we know

$m_2=46\ $\text{kg}$$

$v_2=-4\ $\frac{m}{s}$$ .

After the collision, we have .

Use conservation of momentum to solve this problem.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

$v=\frac{(30\ \text{kg}$)(9\ $\frac{m}{s}$)+(46\ $\text{kg}$)(-4\ $\frac{m}{s}$)}{30\ $\text{kg}$+46\ $\text{kg}$}$

$v=1.13\ $\frac{m}{s}$$

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Question

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

Tap to reveal answer

Answer

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ $\text{kg}$$ ,

Female skater: $m_2=60\ $\text{kg}$$ ,

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_$2v_{2x}$=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_$1v_{1y}$=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{\sin\theta}{\cos\theta}$

$$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\tan\theta$

Plug in the numbers to find the angle

$$\frac{(68\ \text{kg}$)(15\ $\frac{m}{s}$)}{(60\ $\text{kg}$)(12\ $\frac{m}{s}$)}=\tan\theta$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_$2v_{2x}$=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_$2v_{2x}$$}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg}$)(12\ $\frac{m}{s}$)}{(68\ $\text{kg}$+60\ $\text{kg}$)\cos(54.8)}$

$v=9.8\ $\frac{m}{s}$$

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Question

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0$\frac{m}{s}$$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0$\frac{m}{s}$$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

Tap to reveal answer

Answer

Relevant equations:

$K = $$\frac{1}{2}$mv^2$$

$\Delta K = K_f - K_i$

Use conservation of momentum to determine the final velocity of the objects.

$m_$Av_{A,i}$+m_$Bv_{B,i}$=m_$Av_{A,f}$+m_$Bv_{B,f}$$

$m_$Av_{A,i}$+m_$Bv_{B,i}$=(m_A + m_B)v_f$

$(1kg)(6$\frac{m}{s}$)+(5kg)(4$\frac{m}{s}$)=(1kg+5kg)v_f$

$v_f=4.33$\frac{m}{s}$$

Calculate the total initial kinetic energy of the two objects.

$K_${B,i}=\frac{1}{2}$(5kg)(4$\frac{m}{s}$)^2$=40J$

$K_i = $K_{A,i}$$+K_{B,i}$=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_${A,f}=\frac{1}{2}$(1kg)(4.33$\frac{m}{s}$)^2$=9.39J$

$K_${B,f}=\frac{1}{2}$(5kg)(4.33$\frac{m}{s}$)^2$=46.94J$

$K_f = $K_{A,f}$$+K_{B,f}$=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$$\frac{|\Delta K|}{K_i}$100% = $\frac{1.67}{58}$100% \approx 2.9%$

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Question

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

$v_$1^{}$=final\ velocity\ of\ ball\ 1$

$v_$2^{}$=final\ velocity\ of\ ball\ 2$

Tap to reveal answer

Answer

We must use conservation of momentum to tackle this problem.

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ $\text{kg}$$

$v_1=7.1\ $\frac{m}{s}$$

and we know the following for the second ball:

$m_2=0.95\ $\text{kg}$$

$v_2=-2.8\ $\frac{m}{s}$$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

Plug this into our conservation of momentum equation.

So now is eliminated and we can solve for .

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$

$v_2'=\frac{2(0.54\ \text{kg}$)(7.1\ $\frac{m}{s}$)+(0.95\ $\text{kg}$-0.54\ $\text{kg}$)(-2.8\ $\frac{m}{s}$)}{0.54\ $\text{kg}$+0.95\ $\text{kg}$}$

$v_2'=4.4\ $\frac{m}{s}$$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

$v_1'=4.4\ $\frac{m}{s}$-7.1\ $\frac{m}{s}$-2.8\ $\frac{m}{s}$$

$v_1'=-5.5\ $\frac{m}{s}$$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Question

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

Tap to reveal answer

Answer

For the 30 kg cart, we know

$m_1=30\ $\text{kg}$$

$v_1=9\ $\frac{m}{s}$$

and for the 46 g cart, we know

$m_2=46\ $\text{kg}$$

$v_2=-4\ $\frac{m}{s}$$ .

After the collision, we have .

Use conservation of momentum to solve this problem.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

$v=\frac{(30\ \text{kg}$)(9\ $\frac{m}{s}$)+(46\ $\text{kg}$)(-4\ $\frac{m}{s}$)}{30\ $\text{kg}$+46\ $\text{kg}$}$

$v=1.13\ $\frac{m}{s}$$

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Question

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

Tap to reveal answer

Answer

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ $\text{kg}$$ ,

Female skater: $m_2=60\ $\text{kg}$$ ,

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_$2v_{2x}$=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_$1v_{1y}$=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{\sin\theta}{\cos\theta}$

$$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\tan\theta$

Plug in the numbers to find the angle

$$\frac{(68\ \text{kg}$)(15\ $\frac{m}{s}$)}{(60\ $\text{kg}$)(12\ $\frac{m}{s}$)}=\tan\theta$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_$2v_{2x}$=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_$2v_{2x}$$}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg}$)(12\ $\frac{m}{s}$)}{(68\ $\text{kg}$+60\ $\text{kg}$)\cos(54.8)}$

$v=9.8\ $\frac{m}{s}$$

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Question

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0$\frac{m}{s}$$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0$\frac{m}{s}$$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

Tap to reveal answer

Answer

Relevant equations:

$K = $$\frac{1}{2}$mv^2$$

$\Delta K = K_f - K_i$

Use conservation of momentum to determine the final velocity of the objects.

$m_$Av_{A,i}$+m_$Bv_{B,i}$=m_$Av_{A,f}$+m_$Bv_{B,f}$$

$m_$Av_{A,i}$+m_$Bv_{B,i}$=(m_A + m_B)v_f$

$(1kg)(6$\frac{m}{s}$)+(5kg)(4$\frac{m}{s}$)=(1kg+5kg)v_f$

$v_f=4.33$\frac{m}{s}$$

Calculate the total initial kinetic energy of the two objects.

$K_${B,i}=\frac{1}{2}$(5kg)(4$\frac{m}{s}$)^2$=40J$

$K_i = $K_{A,i}$$+K_{B,i}$=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_${A,f}=\frac{1}{2}$(1kg)(4.33$\frac{m}{s}$)^2$=9.39J$

$K_${B,f}=\frac{1}{2}$(5kg)(4.33$\frac{m}{s}$)^2$=46.94J$

$K_f = $K_{A,f}$$+K_{B,f}$=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$$\frac{|\Delta K|}{K_i}$100% = $\frac{1.67}{58}$100% \approx 2.9%$

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Question

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

$v_$1^{}$=final\ velocity\ of\ ball\ 1$

$v_$2^{}$=final\ velocity\ of\ ball\ 2$

Tap to reveal answer

Answer

We must use conservation of momentum to tackle this problem.

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ $\text{kg}$$

$v_1=7.1\ $\frac{m}{s}$$

and we know the following for the second ball:

$m_2=0.95\ $\text{kg}$$

$v_2=-2.8\ $\frac{m}{s}$$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

Plug this into our conservation of momentum equation.

So now is eliminated and we can solve for .

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$

$v_2'=\frac{2(0.54\ \text{kg}$)(7.1\ $\frac{m}{s}$)+(0.95\ $\text{kg}$-0.54\ $\text{kg}$)(-2.8\ $\frac{m}{s}$)}{0.54\ $\text{kg}$+0.95\ $\text{kg}$}$

$v_2'=4.4\ $\frac{m}{s}$$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

$v_1'=4.4\ $\frac{m}{s}$-7.1\ $\frac{m}{s}$-2.8\ $\frac{m}{s}$$

$v_1'=-5.5\ $\frac{m}{s}$$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Question

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

Tap to reveal answer

Answer

For the 30 kg cart, we know

$m_1=30\ $\text{kg}$$

$v_1=9\ $\frac{m}{s}$$

and for the 46 g cart, we know

$m_2=46\ $\text{kg}$$

$v_2=-4\ $\frac{m}{s}$$ .

After the collision, we have .

Use conservation of momentum to solve this problem.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

$v=\frac{(30\ \text{kg}$)(9\ $\frac{m}{s}$)+(46\ $\text{kg}$)(-4\ $\frac{m}{s}$)}{30\ $\text{kg}$+46\ $\text{kg}$}$

$v=1.13\ $\frac{m}{s}$$

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Question

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

Tap to reveal answer

Answer

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ $\text{kg}$$ ,

Female skater: $m_2=60\ $\text{kg}$$ ,

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_$2v_{2x}$=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_$1v_{1y}$=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{\sin\theta}{\cos\theta}$

$$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\tan\theta$

Plug in the numbers to find the angle

$$\frac{(68\ \text{kg}$)(15\ $\frac{m}{s}$)}{(60\ $\text{kg}$)(12\ $\frac{m}{s}$)}=\tan\theta$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_$2v_{2x}$=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_$2v_{2x}$$}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg}$)(12\ $\frac{m}{s}$)}{(68\ $\text{kg}$+60\ $\text{kg}$)\cos(54.8)}$

$v=9.8\ $\frac{m}{s}$$

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Question

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0$\frac{m}{s}$$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0$\frac{m}{s}$$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

Tap to reveal answer

Answer

Relevant equations:

$K = $$\frac{1}{2}$mv^2$$

$\Delta K = K_f - K_i$

Use conservation of momentum to determine the final velocity of the objects.

$m_$Av_{A,i}$+m_$Bv_{B,i}$=m_$Av_{A,f}$+m_$Bv_{B,f}$$

$m_$Av_{A,i}$+m_$Bv_{B,i}$=(m_A + m_B)v_f$

$(1kg)(6$\frac{m}{s}$)+(5kg)(4$\frac{m}{s}$)=(1kg+5kg)v_f$

$v_f=4.33$\frac{m}{s}$$

Calculate the total initial kinetic energy of the two objects.

$K_${B,i}=\frac{1}{2}$(5kg)(4$\frac{m}{s}$)^2$=40J$

$K_i = $K_{A,i}$$+K_{B,i}$=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_${A,f}=\frac{1}{2}$(1kg)(4.33$\frac{m}{s}$)^2$=9.39J$

$K_${B,f}=\frac{1}{2}$(5kg)(4.33$\frac{m}{s}$)^2$=46.94J$

$K_f = $K_{A,f}$$+K_{B,f}$=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$$\frac{|\Delta K|}{K_i}$100% = $\frac{1.67}{58}$100% \approx 2.9%$

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Question

We have two balls. The first ball has mass 0.54kg and is traveling 7.1m/s to the right. It collides head-on elastically with a second ball of mass 0.95kg traveling 2.8m/s to the left. After the collision, what is the speed and direction of each ball?

$v_$1^{}$=final\ velocity\ of\ ball\ 1$

$v_$2^{}$=final\ velocity\ of\ ball\ 2$

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Answer

We must use conservation of momentum to tackle this problem.

We are to find and , the velocities of the balls after the collision. We know the following for the first ball:

$m_1=0.54\ $\text{kg}$$

$v_1=7.1\ $\frac{m}{s}$$

and we know the following for the second ball:

$m_2=0.95\ $\text{kg}$$

$v_2=-2.8\ $\frac{m}{s}$$ .

After plugging these values into our conservation of momentum equation, it is clear that we can't use this equation alone to find and ; however, since we are dealing with an elastic collision, we can use the relation below.

This relation can be derived using conservation of momentum and conservation of kinetic energy equations. Remember that kinetic energy is only conserved if the collision is elastic. We can use this relation to eliminate either or in our conservation of momentum equation.

Plug this into our conservation of momentum equation.

So now is eliminated and we can solve for .

$v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$

$v_2'=\frac{2(0.54\ \text{kg}$)(7.1\ $\frac{m}{s}$)+(0.95\ $\text{kg}$-0.54\ $\text{kg}$)(-2.8\ $\frac{m}{s}$)}{0.54\ $\text{kg}$+0.95\ $\text{kg}$}$

$v_2'=4.4\ $\frac{m}{s}$$

This is the speed of the second ball, and it is traveling to the right because of the positive value. Use this positive value to find by using the relation we found earlier.

$v_1'=4.4\ $\frac{m}{s}$-7.1\ $\frac{m}{s}$-2.8\ $\frac{m}{s}$$

$v_1'=-5.5\ $\frac{m}{s}$$

This is the speed of the first ball and it is traveling to the left due to the negative sign.

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Question

A 30kg cart travels at 9m/s and it hits another cart of mass 46kg traveling at 4m/s in the opposite direction. After the collision, they stick together to form one cart. Find the speed of this cart

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Answer

For the 30 kg cart, we know

$m_1=30\ $\text{kg}$$

$v_1=9\ $\frac{m}{s}$$

and for the 46 g cart, we know

$m_2=46\ $\text{kg}$$

$v_2=-4\ $\frac{m}{s}$$ .

After the collision, we have .

Use conservation of momentum to solve this problem.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

$v=\frac{(30\ \text{kg}$)(9\ $\frac{m}{s}$)+(46\ $\text{kg}$)(-4\ $\frac{m}{s}$)}{30\ $\text{kg}$+46\ $\text{kg}$}$

$v=1.13\ $\frac{m}{s}$$

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Question

There are two skaters. The male skater with mass 68kg travels 15m/s North. He approaches a 60kg female skater who is travel 12m/s East; they approach each other at right angles. When they meet, they hold on to each other. At what direction and speed do they move after they meet?

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Answer

This is a two-dimensional inelastic collision problem and we use conservation of momentum to solve. We know the following.

Male skater: $m_1=68\ $\text{kg}$$ ,

Female skater: $m_2=60\ $\text{kg}$$ ,

First, write down two equations representing conservation of momentum. One equation represents momentum in the x-direction (East-West direction), which is $m_$2v_{2x}$=(m_1+m_2)v\cos\theta$ ,and the other gives the momentum in the y-direction (North-South direction), $m_$1v_{1y}$=(m_1+m_2)v\sin\theta$ .

Take the y-direction of momentum and divide it by the x-direction momentum.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{(m_1+m_2)v\sin\theta}{(m_1+m_2)v\cos\theta}$

Simplify.

$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\frac{\sin\theta}{\cos\theta}$

$$\frac{m_$1v_{1y}$$}{m_$2v_{2x}$}=\tan\theta$

Plug in the numbers to find the angle

$$\frac{(68\ \text{kg}$)(15\ $\frac{m}{s}$)}{(60\ $\text{kg}$)(12\ $\frac{m}{s}$)}=\tan\theta$

To find the speed at which they move at this angle we can use one of the momentum equations.

$m_$2v_{2x}$=(m_1+m_2)v\cos\theta$

Solve for v.

$v=\frac{m_$2v_{2x}$$}{(m_1+m_2)\cos\theta}$

$v=\frac{(60\ \text{kg}$)(12\ $\frac{m}{s}$)}{(68\ $\text{kg}$+60\ $\text{kg}$)\cos(54.8)}$

$v=9.8\ $\frac{m}{s}$$

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Question

Object A has mass $\small 1.0kg$ and initially moves to the right at $\small 6.0$\frac{m}{s}$$ . It then collides with object B, which has mass $\small 5.0kg$ and was initially moving to the right at $\small 4.0$\frac{m}{s}$$ . If the two objects stick together after the collision, what percentage of the initial kinetic energy has been dissipated?

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Answer

Relevant equations:

$K = $$\frac{1}{2}$mv^2$$

$\Delta K = K_f - K_i$

Use conservation of momentum to determine the final velocity of the objects.

$m_$Av_{A,i}$+m_$Bv_{B,i}$=m_$Av_{A,f}$+m_$Bv_{B,f}$$

$m_$Av_{A,i}$+m_$Bv_{B,i}$=(m_A + m_B)v_f$

$(1kg)(6$\frac{m}{s}$)+(5kg)(4$\frac{m}{s}$)=(1kg+5kg)v_f$

$v_f=4.33$\frac{m}{s}$$

Calculate the total initial kinetic energy of the two objects.

$K_${B,i}=\frac{1}{2}$(5kg)(4$\frac{m}{s}$)^2$=40J$

$K_i = $K_{A,i}$$+K_{B,i}$=18J+40J=58J$

Calculate the total final kinetic energy of the two objects.

$K_${A,f}=\frac{1}{2}$(1kg)(4.33$\frac{m}{s}$)^2$=9.39J$

$K_${B,f}=\frac{1}{2}$(5kg)(4.33$\frac{m}{s}$)^2$=46.94J$

$K_f = $K_{A,f}$$+K_{B,f}$=9.39J+46.94J = 56.33J$

Find the difference in kinetic energy.

$\Delta K = K_f-K_i = 56.33J - 58J = -1.67J$

This is the amount of kinetic energy lost during the inelastic collision. Express this amount as a percentage of the initial kinetic energy.

$$\frac{|\Delta K|}{K_i}$100% = $\frac{1.67}{58}$100% \approx 2.9%$

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