Question 1

A baseball is struck by a batter at an angle such that it reaches its maximum height at the location of the pitcher – and just grazes the top of the pitcher’s head (at an elevation of 1.50 m above the launch point of the baseball).  The pitcher is a horizontal distance of 20.0 m from the point at which the ball was struck.  What was the launch speed for the baseball?
A) 37 m/s 
B) 36 m/s 
C) 51 m/s 
D) 27 m/s

Accepted Answer

The maximum height of the projectile is given as 1.50 m. Using the kinematic equations and the fact that the horizontal distance to the maximum height is 20.0 m, we can calculate the launch speed. The correct calculation leads to a launch speed of approximately 37 m/s.

Question 2

A 92.5 kg stunt man stands on a piston that will launch him out of a vertical tube.The piston applies a constant force F to the man's feet.Starting from rest,the force is applied as the man moves vertically 3.25 m,and while the piston is pushing,the man attains a velocity of 25.5 m/s.What was the man's apparent weight during this time interval?
A) 9.3 kN
B) 18.5 kN
C) 19.4 kN
D) 10.2 kN

Accepted Answer

The answer of A 92.5 kg stunt man stands on...

Question 3

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 6.00 s and then at constant velocity for the remaining time,what was his average velocity? A) 8.8 m/s B) 10.2 m/s C) 9.9 m/s D) 9.8 m/s

Accepted Answer

The answer of A sprinter runs 100.0 m.If he travels...

Question 4

On the way to lower floors,an elevator begins its descent from rest at a constant acceleration,descending the first 5.0 m in 0.85 s.What is the apparent weight of a 75 kg man inside the accelerator during this time interval?
A) 0.74 kN
B) 1.8 kN
C) 0.30 kN
D) 1.0 kN

Accepted Answer

Firstly we need to find the acceleration of the elevator:

We know that the elevator starts from rest, so the initial velocity u = 0 m/s

The final velocity v can be found using the equation:

v = u + at

where t = 0.85 s (the time it takes to descend 5.0 m)

v = 0 + a(0.85)
v = 0.85a

We also know that the distance travelled, s = 5.0 m

Using the equation:

s = ut + 1/2at^2

5.0 = 0 + 1/2a(0.85)^2

a = 6.32 m/s^2

Now we can find the apparent weight of the man using the equation:

F = ma

where m = 75 kg (the mass of the man)

F = 75 × 9.81 - 75 × 6.32 (since the acceleration is downwards)

F = 558.75 - 474

F = 84.75 N

Converting this to kN:

F = 0.08475 kN

Therefore, the apparent weight of the man is 0.08475 kN, which is closest to option B (1.8 kN).

Question 5

A model rocket flies upward at constant acceleration,and undergoes a displacement of 15.0 m during its first 0.25 s of travel.What was its speed at the end of the first 0.25 s?
A) 30 m/s
B) 60 m/s
C) 11 m/s
D) 120 m/s

Accepted Answer

The answer of A model rocket flies upward at constant...

Question 6

A small golden statue falls from a height of 8.2 m above the top surface of a pillow.If the acceleration of the statue upon striking the pillow is constant,what is the value of the acceleration if the pillow compresses by 3.9 cm as the statue comes to rest? A) 2100 m/s² B) 21 m/s² C) 160 m/s² D) 210 m/s²

Accepted Answer

The acceleration of the statue is constant, so we can use the equation of motion:v^2 = u^2 + 2aswhere v is the final velocity (0 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled (0.039 m).Substituting these values into the equation, we get:0^2 = 0^2 + 2a(0.039)0 = 0.078aa = 0 m/s^2Therefore, the acceleration of the statue is 0 m/s^2.

Question 7

A 35.0 kg girl swings on a swingset.At the bottom of the swing's path,she feels as though she weighs 435 N.What is the value of the tension in the swing's ropes at the bottom of the swing's path?
A) 92 N
B) 435 N
C) 343 N
D) 778 N

Accepted Answer

The answer of A 35.0 kg girl swings on a...

Question 8

A 92.5 kg stunt man stands on a piston that will launch him out of a vertical tube.The piston applies a constant force F to the man's feet.Starting from rest,the force is applied for 1.20 s,and during that time interval,the man attains a velocity of 25.5 m/s.What was the man's apparent weight during this time interval?
A) Unable to answer without knowing the value of F
B) 3740 N
C) 1970 N
D) 2830 N
E) 2870 N

Accepted Answer

The answer of A 92.5 kg stunt man stands on...

Question 9

If a flea can jump up to 45 cm above its initial resting point,with what speed is it able to leave the ground when jumping?
A) 2.1 m/s
B) 3.0 m/s
C) 8.8 m/s
D) 4.2 m/s

Accepted Answer

The speed can be calculated using the formula for the maximum height of a projectile: $ v^2 = 2gh $, where $ v $ is the initial velocity, $ g $ is the acceleration due to gravity (approximately 9.8 m/s²), and $ h $ is the height (0.45 m). Solving for $ v $, we get $ v = \sqrt{2 \times 9.8 \times 0.45} \approx 3.0 $ m/s.

Question 10

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate of 0.25 m/s² until he reaches his hole safely.How long a time was required for him to cover the last 1.25 m? A) 1.4 s B) 1.7 s C) 1.2 s D) 10 s

Accepted Answer

Using the equation of motion $ s = ut + \frac{1}{2}at^2 $, where $ s = 1.25 \, \text{m} $, $ u = 0.75 \, \text{m/s} $, and $ a = 0.25 \, \text{m/s}^2 $, we solve for $ t $ to find $ t \approx 1.4 \, \text{s} $.

Question 11

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 75.0 m and then at constant velocity for the remaining distance,what was his total elapsed time for the whole race? A) 11.3 s B) 9.8 s C) 10.1 s D) 10.2 s

Accepted Answer

To solve this problem, we can divide it into two parts: the acceleration phase and the constant velocity phase.1. **Acceleration Phase (First 75.0 m):**Given acceleration $a = 1.95 \, \text{m/s}^2$ and distance $s = 75.0 \, \text{m}$, we can use the equation $s = \frac{1}{2} a t^2$ to find the time $t$ it takes to cover this distance. Rearranging the equation for $t$, we get $t = \sqrt{\frac{2s}{a}}$.Plugging in the given values: $t = \sqrt{\frac{2 \times 75.0}{1.95}} = \sqrt{\frac{150}{1.95}} \approx 8.7 \, \text{s}$.2. **Constant Velocity Phase (Last 25.0 m):**To find the velocity at the end of the acceleration phase, we use $v = at$, where $v$ is the final velocity and $t$ is the time found in the acceleration phase.$v = 1.95 \times 8.7 \approx 16.965 \, \text{m/s}$.This velocity is maintained for the remaining 25.0 m. The time to cover this distance at constant velocity is $t = \frac{s}{v} = \frac{25.0}{16.965} \approx 1.47 \, \text{s}$.3. **Total Time:**Adding the times from both phases gives the total time: $8.7 \, \text{s} + 1.47 \, \text{s} \approx 10.17 \, \text{s}$, which rounds to $10.2 \, \text{s}$, making option D the correct answer.

Question 12

One stone is dropped from the side of a bridge,and a second is dropped from the same place a short time later.What is true about the distance between the stones as a function of time after both have been dropped? Ignore air resistance
A) Increases,finally reaching a constant value
B) Always increases
C) Is constant
D) Decreases,and then reaches a constant value
E) Always decreases

Accepted Answer

The distance between the stones always increases because the first stone has been falling for a longer time and thus has a greater velocity than the second stone, causing the gap between them to widen over time.

Question 13

A skydiver has bailed out of his airplane at a height of 3000 m.The mass of the skydiver and his parachute is 80 kg.What is the drag (force of air resistance)when he reaches terminal speed?
A) 120 N
B) 360 N
C) 784
D) 750 N

Accepted Answer

At terminal speed, the drag force equals the gravitational force. The gravitational force is calculated as mass (80 kg) times gravity (9.8 m/s²), which equals 784 N.

Question 14

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?
A) 17.7 m/s.
B) 15.4 m/s.
C) 20.3 m/s.
D) 17.5 m/s.

Accepted Answer

The sprinter accelerates for 75 m and then travels at constant velocity for 25 m. Using the total time and distance, we can calculate the peak velocity during the constant velocity phase. The peak velocity is 17.7 m/s.

Question 15

A ball is dropped at time t = 0.0 s.At t = 2.0 s,a second ball is thrown downward with speed v.What is v if at t = 4.0 s,the two balls are at the same vertical position?
A) 15 m/s
B) 25 m/s
C) 49 m/s
D) 29 m/s

Accepted Answer

The answer of A ball is dropped at time t...

Question 16

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?
A) 17.7 m/s.
B) 17.5 m/s.
C) 20.3 m/s.
D) 15.4 m/s.

Accepted Answer

To find the peak velocity, we first calculate the time taken to cover the last 25 m at constant velocity. The total time is 9.87 s, and the time for the first 75 m can be found using the equation for constant acceleration. Solving these equations gives a peak velocity of approximately 17.7 m/s.

Question 17

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate until he reaches his hole safely,0.75 s later.What acceleration did he undergo during the final 1.25 m of his escape? A) 1.8 m/s² B) 1.2 m/s² C) 4.8 m/s² D) 2.4 m/s² E) 6.4 m/s²

Accepted Answer

We can use the following equation to solve the problem:

Δx = ut + 0.5at^2

where:

Δx = displacement (1.25 m) 
u = initial velocity (0.75 m/s) 
t = time (0.75 s) 
a = acceleration (unknown)

Rearranging the equation:

a = (2Δx - 2ut) / t^2

Substituting the values:

a = (2(1.25) - 2(0.75)(0.75)) / (0.75)^2

a = 2.4 m/s^2

Therefore, the mouse underwent an acceleration of 2.4 m/s^2 during the final 1.25 m of its escape.

Quiz 4: Motion With Constant Acceleration

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 6.00 s and then at constant velocity for the remaining time,what was his average velocity?

On the way to lower floors,an elevator begins its descent from rest at a constant acceleration,descending the first 5.0 m in 0.85 s.What is the apparent weight of a 75 kg man inside the accelerator during this time interval?

A model rocket flies upward at constant acceleration,and undergoes a displacement of 15.0 m during its first 0.25 s of travel.What was its speed at the end of the first 0.25 s?

A small golden statue falls from a height of 8.2 m above the top surface of a pillow.If the acceleration of the statue upon striking the pillow is constant,what is the value of the acceleration if the pillow compresses by 3.9 cm as the statue comes to rest?

A 35.0 kg girl swings on a swingset.At the bottom of the swing's path,she feels as though she weighs 435 N.What is the value of the tension in the swing's ropes at the bottom of the swing's path?

If a flea can jump up to 45 cm above its initial resting point,with what speed is it able to leave the ground when jumping?

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate of 0.25 m/s² until he reaches his hole safely.How long a time was required for him to cover the last 1.25 m?

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 75.0 m and then at constant velocity for the remaining distance,what was his total elapsed time for the whole race?

One stone is dropped from the side of a bridge,and a second is dropped from the same place a short time later.What is true about the distance between the stones as a function of time after both have been dropped? Ignore air resistance

A skydiver has bailed out of his airplane at a height of 3000 m.The mass of the skydiver and his parachute is 80 kg.What is the drag (force of air resistance) when he reaches terminal speed?

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?

A ball is dropped at time t = 0.0 s.At t = 2.0 s,a second ball is thrown downward with speed v.What is v if at t = 4.0 s,the two balls are at the same vertical position?

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate until he reaches his hole safely,0.75 s later.What acceleration did he undergo during the final 1.25 m of his escape?

Quiz 4: Motion With Constant Acceleration

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s2 for the first 6.00 s and then at constant velocity for the remaining time,what was his average velocity?

On the way to lower floors,an elevator begins its descent from rest at a constant acceleration,descending the first 5.0 m in 0.85 s.What is the apparent weight of a 75 kg man inside the accelerator during this time interval?

A model rocket flies upward at constant acceleration,and undergoes a displacement of 15.0 m during its first 0.25 s of travel.What was its speed at the end of the first 0.25 s?

A small golden statue falls from a height of 8.2 m above the top surface of a pillow.If the acceleration of the statue upon striking the pillow is constant,what is the value of the acceleration if the pillow compresses by 3.9 cm as the statue comes to rest?

A 35.0 kg girl swings on a swingset.At the bottom of the swing's path,she feels as though she weighs 435 N.What is the value of the tension in the swing's ropes at the bottom of the swing's path?

If a flea can jump up to 45 cm above its initial resting point,with what speed is it able to leave the ground when jumping?

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate of 0.25 m/s2 until he reaches his hole safely.How long a time was required for him to cover the last 1.25 m?

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s2 for the first 75.0 m and then at constant velocity for the remaining distance,what was his total elapsed time for the whole race?

One stone is dropped from the side of a bridge,and a second is dropped from the same place a short time later.What is true about the distance between the stones as a function of time after both have been dropped? Ignore air resistance

A skydiver has bailed out of his airplane at a height of 3000 m.The mass of the skydiver and his parachute is 80 kg.What is the drag (force of air resistance) when he reaches terminal speed?

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?

A ball is dropped at time t = 0.0 s.At t = 2.0 s,a second ball is thrown downward with speed v.What is v if at t = 4.0 s,the two balls are at the same vertical position?

A sprinter runs 100.0 m in 9.87 seconds.If he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m,what was his peak velocity?

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate until he reaches his hole safely,0.75 s later.What acceleration did he undergo during the final 1.25 m of his escape?

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 6.00 s and then at constant velocity for the remaining time,what was his average velocity?

A mouse is running away from a cat at 0.75 m/s and,at 1.25 m from his mouse hole,he accelerates at a constant rate of 0.25 m/s² until he reaches his hole safely.How long a time was required for him to cover the last 1.25 m?

A sprinter runs 100.0 m.If he travels at constant acceleration of 1.95 m/s² for the first 75.0 m and then at constant velocity for the remaining distance,what was his total elapsed time for the whole race?