MCAT Physical - Motion in One Dimension

Suppose a ball with mass of 10kg was dropped from rest from the top of a cliff that is 80m tall. How long will it take the ball to reach the bottom of the cliff?

10s

16s

Explanation

The question tells us that the initial velocity of the ball (v0) is zero and the height (d) of the cliff is 80m. Acceleration (gravity) is 10m/s2. Using the kinematics eqation $d=v_0 t+(1/2)at^{2}$ we can solve for time.

80 = 0 + (1/2)(10)t2

80/5 = t2 = 16

t = 4s

Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.

Over the course of the afternoon, they collide many times. Four collisions are described below.

Collision 1:

Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 5m/s and subsequently slides into a stationary child 2. They remain linked together after the collision.

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at 10m/s, slides into child 1, moving at 2m/s.

Collision 3:

The two children collide while traveling in opposite directions at 10m/s each.

Collision 4:

The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of +8m/s.

Child 1 hooks up a jet motor to a sled, and brings it to the icy lake. She sets up her sled, and launches from a velocity of 8m/s with an acceleration of 15m/s2. After fifteen seconds, how far has she traveled?

1.8km

1800km

1.8m

600km

600m

Explanation

For this question, we have to use a translational motion equation.

A child throws a ball straight up into the air. He throws the ball with an initial velocity of $\small 8\frac{m}{s}$ . Assume there is no air resistance.

How long will it take for the ball to reach a velocity of $\small 0\frac{m}{s}$ , while in the air?

The baseball will never have a velocity of $\small 0\frac{m}{s}$ .

Explanation

In order to solve for the time at which the ball has a velocity of $\small 0\frac{m}{s}$ , we need to use an equation which incorporates all of the known variables. We know the acceleration due to gravity, the initial velocity, and the final velocity. As a result, the best equation to use is the one that allows us to simply solve for the unknown variable, time.

$\small v = v_{0} + at$

$\small 0\frac{m}{s} = 8\frac{m}{s} + (-10\frac{m}{s^{2}})(t)$

$\small t = 0.8 s$

Keep in mind that 1.6s is also a time at which the velocity is $\small 0\frac{m}{s}$ , but this time refers to the point when the boy catches the ball. The question asks for the specific time while the ball is still in the air, so the correct answer, 0.8s, refers to the point where the ball is at its peak height.

If an object is dropped from a height of 450 meters above Earth, what is its velocity just before impact?

67m/s

6.7m/s

95m/s

45m/s

Explanation

First calculate the time it takes to hit the ground using the equation $d=v_o+\frac{1}{2}at^2$ .

We can plug in values (including acceleration due to gravity) and solve for t.

$450m=0\frac{m}{s}+\frac{1}{2}(10\frac{m}{s^2})t^2$

t = 9.49s

Next, find the final velocity with the equation .

$v_f=0\frac{m}{s}+(10\frac{m}{s})(9.49s)=95\frac{m}{s}$

An object is shot upward from the ground at a velocity of 24m/s. How long before it hits the ground?

4.8s

12.0s

9.6s

2.4s

Explanation

First find the time it takes to reach the top of its path using the equation , where the final velocity is 0m/s, initial velocity is 24m/s, and a is the acceleration due to gravity. Solve for time (t).

0 = 24 +(-10)t

t = 2.4s

Notice that this is only the time it takes the object to reach the top of its path, and the question asks for the time it takes the object to reach the ground. In order to find how long it takes to reach the ground simply double this time.

2.4s * 2 = 4.8 s

Which of the following is not a vector quantity?

Kinetic energy

Weight

Momentum

Displacement

Velocity

Explanation

Kinetic energy is calculated by squaring velocity (which we know is a vector). This eliminates its vector properties making kinetic energy a scalar value. All other answer choices are vector quantities.

If an ball is dropped from a cliff $\small 80m$ high, how long will it take the ball to strike the ground?

Explanation

Since acceleration is constant, we can use the appropriate kinematics equation to solve:

$d=v_0t + \frac{1}{2}at^2$

We are given the height of the cliff, which will be equal to the distance traveled. The initial velocity is zero since the ball starts from rest and the acceleration will be equal to the acceleration due to gravity. Use these values to calculate the time.

$80m=(0\frac{m}{s})t+\frac{1}{2}(9.8\frac{m}{s^2})t^2$

$t^2=\frac{80m}{\frac{1}{2}(9.8\frac{m}{s^2})}=16.3s^2$

$t=\sqrt{16.3s^2}=4.04s$

A sprinter running a race accelerates constantly at $0.5\frac{m}{s^2}$ from rest. What is his approximate final velocity as he crosses the finish line?

$10\frac{m}{s}$

$7\frac{m}{s}$

$50\frac{m}{s}$

$14\frac{m}{s}$

Explanation

To answer this question, we must have a solid understanding of the kinematics equations. For this question, we must use an equation relating final velocity, distance, and acceleration.

The best fit for this is $u_{f}^{2} - u_{o}^{2} = 2ax$ .

Since we are solving for final velocity, and we started from rest, we can simplify the equation.

$u_o=0\frac{m}{s}, a=0.5\frac{m}{s^2}, x=100m$

$u_{f}^{2} - (0\frac{m}{s})^{2} = 2(0.5\frac{m}{s^2})(100m)$

$u_{f} = \sqrt{2(0.5\frac{m}{s^2})(100m)} = \sqrt{100\frac{m^2}{s^2}}=10\frac{m}{s}$

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

125m

100m

50m

250m

75m

Explanation

Using the equation we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, , thus we have . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that = 125m.

If a ball was thrown straight up at 10m/s and strikes the ground after two seconds, what maximum height did the ball reach?

10m

15m

Explanation

This question can be solved using multiple strategies. One strategy uses the equation .

vf = 0m/s

vo = 10m/s

a = -10m/s2

We use a negative acceleration because gravity is in the opposite direction of the movement of the ball. When we plug in all the values we find that d= 5m.

Another strategy uses the projectile time. In projectile motion we know that velocity is zero at the maximum height. Using only half of the projectile time, we can solve for the maximum height.

$d = \frac{1}{2}at^2=\frac{1}{2}(10\frac{m}{s^s})(1s)^2=5m$

Motion in One Dimension

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