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Example Questions

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Example Question #13 : Linear Motion

Two balls, one with mass 2kg and one with mass 4kg , are dropped from 30m above the ground. How long does it take the 4kg ball to hit the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

1.27s

2.47s

6.12s

31.23s

17.15s

Correct answer:

2.47s

Explanation:

The equation for the vertical motion for an object in freefall is:

$\Delta y=\frac{1}{2}at^2$

Notice that there is no place for mass anywhere in this equation. This means that the two balls will be in the air for the same amount of time. We simply need to use the distance and acceleration to solve for the time.

Remember that even though the height is 30m , the DISPLACEMENT will be -30m . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$-30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$3.06s^2=\frac{1}{2}t^2$

2*3.06s^2=t^2

6.12s^2=t^2

$\sqrt{6.12s^2}=t^2$

2.47s=t

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Example Question #11 : Motion And Mechanics

Two balls, one with mass 2kg and one with mass 4kg , are dropped from 30m above the ground. How long does it take the 2kg ball to hit the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

6.12s

2.47s

31.23s

1.27s

9.8s

Correct answer:

2.47s

Explanation:

The equation for the vertical motion for an object in freefall is:

$\Delta y=\frac{1}{2}at^2$

Remember that even though the height is 30m , the DISPLACEMENT will be -30m . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$-30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$3.06s^2=\frac{1}{2}t^2$

2*3.06s^2=t^2

6.12s^2=t^2

$\sqrt{6.12s^2}=t^2$

2.47s=t

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Example Question #15 : Linear Motion

Two balls, one with mass 2kg and one with mass 4kg , are dropped from rest. It takes 18s for them to hit the ground. What is the final velocity of the 2kg ball right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$-9.8\frac{m}{s}$

$-176.4\frac{m}{s}$

$-4.9\frac{m}{s}$

$-2\frac{m}{s}$

$-19.6\frac{m}{s}$

Correct answer:

$-176.4\frac{m}{s}$

Explanation:

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

v_f=v_i+at

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for 18s .

Use the values given in the question to solve for the final velocity.

v_f=v_i+at

$v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

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Example Question #16 : Linear Motion

Two balls, one with mass 2kg and one with mass 4kg , are dropped from rest. It takes 18s for them to hit the ground. What is the final velocity of the 4kg ball right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$-9.8\frac{m}{s}$

$-176.4\frac{m}{s}$

$-352.8\frac{m}{s}$

$-39.2\frac{m}{s}$

$-88.2\frac{m}{s}$

Correct answer:

$-176.4\frac{m}{s}$

Explanation:

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

v_f=v_i+at

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for 18s .

Use the values given in the question to solve for the final velocity.

v_f=v_i+at

$v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

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Example Question #11 : Motion And Mechanics

A 7kg crate slides along a frictionless surface. If it maintains a constant velocity of $3\frac{m}{s}$ , what is the net acceleration on the crate?

Possible Answers:

$60\frac{m}{s^2}$

$3\frac{m}{s^2}$

$0\frac{m}{s^2}$

$6.67\frac{m}{s^2}$

$20\frac{m}{s^2}$

Correct answer:

$0\frac{m}{s^2}$

Explanation:

Acceleration is the change in velocity per unit time.

$a=\frac{v_2-v_1}{t}$

Since the velocity does not change from one moment to the next, then there must be no net acceleration on the object.

$a=\frac{3\frac{m}{s}-3\frac{m}{s}}{t}=\frac{0\frac{m}{s}}{t}$

$a=0\frac{m}{s^2}$

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Example Question #11 : Motion And Mechanics

A crate has an initial velocity of $9\frac{m}{s}$ . If it accelerates at a constant rate of $0.5\frac{m}{s^2}$ for 1.5s , what is the final velocity?

Possible Answers:

$0.75\frac{m}{s}$

$5.25\frac{m}{s}$

$1.5\frac{m}{s}$

$8.25\frac{m}{s}$

$9.75\frac{m}{s}$

Correct answer:

$9.75\frac{m}{s}$

Explanation:

For this problem, we use the formula v_f=v_i+at .

We are given the initial velocity, acceleration, and time. Using these values, we can solve for the final velocity.

$v_f=9\frac{m}{s}+(0.5\frac{m}{s^2}*1.5s)$

$v_f=9\frac{m}{s}+0.75\frac{m}{s}$

$v_f=9.75\frac{m}{s}$

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Example Question #19 : Linear Motion

A box moves at $0.11\frac{m}{s}$ across a frictionless surface. Assuming no other forces, how far will it travel in ?

Possible Answers:

0.11m

3.3m

0.037m

33m

0.33m

Correct answer:

0.33m

Explanation:

The relationship between distance, velocity, and time is given by the formula for velocity:

$v=\frac{\Delta x}{\Delta t}$

We can rearrange this equation to solve for the distance.

$\Delta x=v*\Delta t$

Using the values given for the velocity and the time, we can find the distance traveled.

$\Delta x=0.11\frac{m}{s}*3s$

$\Delta x =0.33m$

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Example Question #20 : Linear Motion

A ball rolls at $2.22\frac{m}{s}$ . Assuming no outside forces, how long will it take for the ball to roll 12m ?

Possible Answers:

0.185s

26.64s

16.64s

9.91s

5.41s

Correct answer:

5.41s

Explanation:

The relationship between distance, velocity, and time is given by the equation for velocity:

$v=\frac{\Delta x}{\Delta t}$

We can rearrange this formula for solve for the time.

$\Delta t =\frac{\Delta x}{v}$

Using the given values for the velocity and distance traveled, we can solve for the time.

$\Delta t =\frac{12m}{2.22\frac{m}{s}}$

$\Delta t=5.41s$

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Example Question #21 : Understanding Motion In One Dimension

A ball rolls at $0.77\frac{m}{s}$ . Assuming no outside forces, how long will it take to roll 9.23m ?

Possible Answers:

11.99s

5.82s

0.141s

7.12s

0.083s

Correct answer:

11.99s

Explanation:

The relationship between distance, velocity, and time is given by the equation for velocity:

$v=\frac{\Delta x}{\Delta t}$

We can rearrange this formula for solve for the time.

$\Delta t =\frac{\Delta x}{v}$

Using the given values for the velocity and distance traveled, we can solve for the time.

$\Delta t =\frac{9.23m}{0.77\frac{m}{s}}$

$\Delta t=11.99s$

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Example Question #21 : Linear Motion

A ball begins to roll with a velocity of $0.02\frac{m}{s}$ . If it has an acceleration of $0.5\frac{m}{s^2}$ and rolls for , what is the final velocity?

Possible Answers:

$0.42\frac{m}{s}$

$8.04\frac{m}{s}$

$2.01\frac{m}{s}$

$0.84\frac{m}{s}$

$4.02\frac{m}{s}$

Correct answer:

$4.02\frac{m}{s}$

Explanation:

For this question, we will need to use the acceleration formula:

$a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{\Delta t}$

We are given the initial velocity, acceleration, and time. Using these values in the equation, we can solve for the final velocity.

$0.5\frac{m}{s^2}=\frac{v_2-0.02\frac{m}{s}}{8s}$

$4\frac{m}{s}=v_2-0.02\frac{m}{s}$

$4.02\frac{m}{s}=v_2$

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #13 : Linear Motion

Example Question #11 : Motion And Mechanics

Example Question #15 : Linear Motion

Example Question #16 : Linear Motion

Example Question #11 : Motion And Mechanics

Example Question #11 : Motion And Mechanics

Example Question #19 : Linear Motion

Example Question #20 : Linear Motion

Example Question #21 : Understanding Motion In One Dimension

Example Question #21 : Linear Motion

All High School Physics Resources

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