Linear Motion and Momentum

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AP Physics 1 › Linear Motion and Momentum

Questions 1 - 10

A blue rubber ball weighing is rolling with a velocity of $2\frac{m}{s}$ when it hits a still red rubber ball with a weight of . After this elastic collision, what are the speeds and directions of the blue and red balls respectively?

$\frac{2}{3} \frac{m}{s}$ to the left, $\frac{4}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the left, $\frac{2}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the right, $\frac{2}{3} \frac{m}{s}$ to the left

At rest, $2 \frac{m}{s}$ to the right

At rest, $1 \frac{m}{s}$ to the right

Explanation

Because we are solving for two velocities (two unknowns), we need two equations. We can use the conservation of linear momentum:

$m_{1}v_{1}+m_{2}v_{2} = m_{1}v_{1}'+m_{2}v_{2}'$

Because we know that the collision is elastic, we know that kinetic energy is conserved:

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$

The red ball starts at rest so

$v_{2}=0$

The above equations can then be simplified and one can solve for $v_{1}'$ and $v_{2}'$ .

$v_{1}'=\left ( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{20-40}{20+40} \right )2=-\frac{2}{3}$

Negative implies the ball is moving to the left.

$v_{2}'=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{2(20)}{20+40} \right )2=\frac{4}{3}$

A object is moving along with the velocity given below. Calculate the magnitude of the momentum vector $\vec{p}$ .

$\vec{v}=5\hat{i}-2\hat{j}$

$162 kg\frac{m}{s}$

$90kg\frac{m}{s}$

$210kg\frac{m}{s}$

$58kg\frac{m}{s}$

$300kg\frac{m}{s}$

Explanation

We begin by writing down the definition of an object's linear momentum $\vec{p}$

$\vec{p}=m\vec{v}$

$\vec p=30kg\left(5\hat{i}-2\hat{j} \right )\frac{m}{s}$

$\vec p=\left( 150\hat{i}-60\hat{j}\right )kg\frac{m}{s}$

We then find the magnitude of the momentum by taking the square root of the sum of squares of its components.

$\left| \vec{p}\right| = \sqrt{p_x^2+p_y^2}$

$\left|\vec p\right|=\sqrt{150^2+60^2}kg\frac{m}{s}=162kg\frac{m}{s}$

A bullet weighing is fired at a velocity of $360\frac{m}{s}$ at a block weighing at rest on a frictionless surface. When the bullet hits the block, it becomes lodged in the block and causes the block to move. How fast does the block move after the collision?

$20.4 \frac{m}{s}$

$354 \frac{m}{s}$

$21.6\frac{m}{s}$

$4.52\frac{m}{s}$

$6.0\frac{m}{s}$

Explanation

Using the equation of momentum,

we can compare the initial and final scenarios and set them equal to each other to solve for the final velocity of the block (don't forget to convert the units of to !):

$m_{1}v_{1}=m_{2}v_{2}$

$.06\times360 = (.06+1)\times v_{2}$

$v_2\approx 20.4$

A object is moving along with the velocity given below. Calculate the magnitude of the momentum vector $\vec{p}$ .

$\vec{v}=5\hat{i}-2\hat{j}$

$162 kg\frac{m}{s}$

$90kg\frac{m}{s}$

$210kg\frac{m}{s}$

$58kg\frac{m}{s}$

$300kg\frac{m}{s}$

Explanation

We begin by writing down the definition of an object's linear momentum $\vec{p}$

$\vec{p}=m\vec{v}$

$\vec p=30kg\left(5\hat{i}-2\hat{j} \right )\frac{m}{s}$

$\vec p=\left( 150\hat{i}-60\hat{j}\right )kg\frac{m}{s}$

We then find the magnitude of the momentum by taking the square root of the sum of squares of its components.

$\left| \vec{p}\right| = \sqrt{p_x^2+p_y^2}$

$\left|\vec p\right|=\sqrt{150^2+60^2}kg\frac{m}{s}=162kg\frac{m}{s}$

$\frac{2}{3} \frac{m}{s}$ to the left, $\frac{4}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the left, $\frac{2}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the right, $\frac{2}{3} \frac{m}{s}$ to the left

At rest, $2 \frac{m}{s}$ to the right

At rest, $1 \frac{m}{s}$ to the right

Explanation

Because we are solving for two velocities (two unknowns), we need two equations. We can use the conservation of linear momentum:

$m_{1}v_{1}+m_{2}v_{2} = m_{1}v_{1}'+m_{2}v_{2}'$

Because we know that the collision is elastic, we know that kinetic energy is conserved:

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$

The red ball starts at rest so

$v_{2}=0$

The above equations can then be simplified and one can solve for $v_{1}'$ and $v_{2}'$ .

$v_{1}'=\left ( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{20-40}{20+40} \right )2=-\frac{2}{3}$

Negative implies the ball is moving to the left.

$v_{2}'=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{2(20)}{20+40} \right )2=\frac{4}{3}$

$20.4 \frac{m}{s}$

$354 \frac{m}{s}$

$21.6\frac{m}{s}$

$4.52\frac{m}{s}$

$6.0\frac{m}{s}$

Explanation

Using the equation of momentum,

we can compare the initial and final scenarios and set them equal to each other to solve for the final velocity of the block (don't forget to convert the units of to !):

$m_{1}v_{1}=m_{2}v_{2}$

$.06\times360 = (.06+1)\times v_{2}$

$v_2\approx 20.4$

Given vector $\vec{A}$ has a magnitude of directed $35^{\circ}$ above the $+\hat{x}$ axis, and vector $\vec{B}$ has a magnitude of directed $70^{\circ}$ above the $+\hat{x}$ axis, calculate $\vec{A}\cdot \vec{B}$ .

Explanation

By definition, the dot product of two vectors can be related to their magnitudes and the angle between them as follows:

$\vec{A}\cdot\vec{B}=\left| \vec{A}\right| \left| \vec{B}\right|\cos{\theta}$

Given the angle between the two vectors is $\theta = 75^{\circ}-35^{\circ}=35^{\circ}$ , we can calculate the dot product to be written explicitly as:

$\vec{A}\cdot\vec{B}=5m \left( 8m\right)\cos{35^{\circ}}=32.8m^2$

Note that the unts are since the dot product involves multiplying two meters together.

Explanation

By definition, the dot product of two vectors can be related to their magnitudes and the angle between them as follows:

$\vec{A}\cdot\vec{B}=\left| \vec{A}\right| \left| \vec{B}\right|\cos{\theta}$

Given the angle between the two vectors is $\theta = 75^{\circ}-35^{\circ}=35^{\circ}$ , we can calculate the dot product to be written explicitly as:

$\vec{A}\cdot\vec{B}=5m \left( 8m\right)\cos{35^{\circ}}=32.8m^2$

Note that the unts are since the dot product involves multiplying two meters together.

A hungry wasp spots an fly wandering about. Assuming the wasp attacks the fly from behind (they are both traveling in the same direction) with speed v, and the fly is stationary, what is the speed of the wasp and fly after the collision? Assume the fly and wasp are one object after the collision. Your answer should be in terms of M, m, v where M is the mass of the wasp, m is the mass of the fly and v is the original speed of the wasp.

$\frac{Mv}{M+m}$

$\frac{mv}{M}$

$\frac{M+m}{Mv}$

$\frac{mv}{M+m}$

, they are both stationary after the collision.

Explanation

Considering the wasp aims to eat the fly, we assume the fly and wasp are one body after the collision. This is an inelastic collision. We can solve this with conservation of momentum.

$\Delta P=0$ or $P_{f}=P_{i}$

For the two body inelastic colision between the wasp and the fly, we can rewrite this as:

$(mv)_{wasp}+(mv)_{fly}=(m_{wasp}+m_{fly})v_{fly+wasp}$

Then taking into account the fact the fly is stationary initially:

$Mv+0=(M+m)v_{fly+wasp}$

Then solve for the velocity of the fly and the wasp after the collision:

$v_{fly+wasp}=\frac{Mv}{M+m}$

One car with a mass of 400kg is traveling east at $20\frac{m}{s}$ and collides with a car of mass 800kg traveling west at $15\frac{m}{s}$ . Assuming the collision is completely inellastic, what is velocity of the first car after the collision?

$3.33\frac{m}{s}\ \text{west}$

$3.33\frac{m}{s}\ \text{east}$

$1.0\frac{m}{s}\ \text{west}$

$1.0\frac{m}{s}\ \text{east}$

$0.0\frac{m}{s}$

Explanation

Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:

There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:

Rearranging for final velocity, we get:

$v = \frac{m_1v_1 + m_2v_2}{m}$

At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:

$v = \frac{(400)(-20)+ (800)(15)}{400 + 800}$