Motion in One Dimension - MCAT Physical

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Question

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?

Answer

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

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Question

A car is moving with a constant velocity of $10\frac{m}{s}$ when it abruptly stops, applying a constant breaking acceleration of $-5\frac{m}{s^2}$ . Over what distance does the car come to a stop?

Answer

This problem can be easily solved using the formula, $v{_{f}}^{2} = v{_{0}}^{2} + 2ax$ , and solving for . Our initial velocity is $10\frac{m}{s}$ , acceleration is $-5\frac{m}{s^2}$ , and final velocity is $0\frac{m}{s}$ .

$(0\frac{m}{s})^2=(10\frac{m}{s})^2+2(-5\frac{m}{s^2})x$

$0=100\frac{m^2}{s^2}-10\frac{m}{s^2}x$

$-100\frac{m^2}{s^2}=-10\frac{m}{s^2}x$

$\frac{-100\frac{m^2}{s^2}}{-10\frac{m}{s^2}}=10m=x$

The car comes to a stop after applying the breaks over 10m.

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Question

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

Answer

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 .

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

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Question

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

Answer

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

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Question

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 +8 m/s with an acceleration of 15m/s2. After ten seconds, how fast is she traveling? Assume the lake is frictionless.

Answer

For this question, we have to use the translational motion equations:

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Question

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?

Answer

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

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Question

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

Answer

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.

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Question

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

Answer

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.

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

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

The rope is cut right above the 2 kgmass. What is the acceleration of the platform with the individual weights as it falls to the ground?

Answer

Remember that Newton’s second law is applicable regardless of whether we are talking about kinematics, torque, electric, magnetic, or gravitational force, thus, F = ma. We know that the only force acting in free-fall in an environment where we can neglect air resistance (i.e. most of the problems on the MCAT) is gravity. The acceleration due to gravity is 9.8m/s2. The sign could be + or – depending on how you decided to orient.

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the velocity of the platform two seconds after the rope is cut?

Answer

Think back to the 3 main kinematics equations that we know.

vf2 = vi2 + 2aΔx

vf = vi + at

Δx = vit + ½at2

We need to determine which formula will allow us to find final velocity after a given amount of time.

vf = vi + at = (0m/s) + (9.8m/s2)(2s) = 19.6m/s

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Question

If a ball is thrown straight up into the air at a velocity of 20m/s and strikes the ground five seconds later, what is the velocity of the ball at its maximum height?

Answer

The velocity of the ball at the maximum height is equal to zero. Since the ball was thrown stright up, both the velocity in the x direction and the y direction are equal to zero at its peak. If the ball were thrown at an angle, only the velocity in the y direction would be equal to zero. This is a good point to keep in mind when doing kinematic problems. Note that there is still acceleration acting on the ball (gravity) of 10m/s2 which brings the ball back to the ground.

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Question

If an object has an initial velocity of 3m/s and a constant acceleration of 2m/s2, how fast will the object be moving after five seconds?

Answer

Using the equation we can find the final velocity of the object after five seconds.

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Question

How far will an object travel after ten seconds if it is dropped into a bottomless pit?

Answer

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation can be used to find the distance traveled.

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Question

In five seconds a runner moves from the 50m mark to the 75m mark along a straight path. What is the runner's average velocity?

Answer

To find the average velocity, the displacement needs to be measured. The displacement the difference betweent eh final position and the initial position.

Using the equation for velocity, $velocity=\frac{displacement}{time}$ , we can find the average velocity.

$v=\frac{25m}{5s}=5\frac{m}{s}$

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Question

How long does it take an object to travel a distance of 30m from rest at a constant acceleration of 2m/s2?

Answer

Using the equation , we can solve for time.

Since the object started at rest, . Now we are left with the equation .

$30m=\frac{1}{2}(2\frac{m}{s^2})t^2$

Plugging in the remaining values we can find that t = 5.5s.

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Question

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?

Answer

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.

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Question

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

Answer

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.

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

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

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

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Question

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?

Answer

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.

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

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

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

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Question

An car accelerates from rest and travels in . What is the approximate acceleration of the car?

Answer

The kinematics equation relating distance, time, and acceleration is $x = \upsilon_{o}t + \frac{1}{2}at^{2}$ .

Since we know that the car was initially at rest, we can rewrite the equation with zero initial velocity.

$x=\frac{1}{2}at^2$

Now we can plug in our values from the question, and solve for the acceleration.

$25m=\frac{1}{2}a(0.7s)^2$

$25m=\frac{1}{2}a(0.49s^2)=(0.245s^2)a$

$a=\frac{25m}{0.245s^2}=102\frac{m}{s^2}\approx 100\frac{m}{s^2}$

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Question

A ball is dropped off of a roof. If the ball takes four seconds to hit the ground, what is the ball's speed right before it hits the ground?

Answer

We are given the time that the ball is in the air and the acceleration. Since the ball is dropped, we can also tell that the initial velocity is zero (starts from rest). Using these given values, we can calculate the final velocity using the appropriate motion equation.

$v_f = 0\frac{m}{s} + (-9.8\frac{m}{s^2})(4s)= -39.2\frac{m}{s}$

The final velocity is negative because the ball is traveling downward; however, the question asks for speed, allowing us to disregard the vector indicator. Speed cannot be negative.

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