PHY 111 FINAL EXAM QUESTIONS
AND ANSWERS 2026
A toy car with mass m1 (0.36 kg ) travels to the right on a frictionless track with a speed
of 3 m/s. A second toy car with a mass m2 (0.74 kg) travels with a speed of 6 m/s to the
left on the same track. The two cars collide and stick together. What are the cars'
speeds after the collision? - ANSWERSmomentum conservation eq:
m1v1 - m2v2 = (m1 + m2)v
where: m1 = 0.36 kg, v1 = 3 m/s, m2 = 0.74 kig, v2 = 6 m/s
(0.36)(3) - (0.74)(6) = (0.36 + 0.74)v
v = 3.05 m/s
A student fires a 0.07 kg arrow at an object with mass m (6.1 kg) that is initially at rest
on a frictionless surface. The speed of the arrow before the collision is 90 m/s. The
speed when the arrow emerges from the object is v (0.78 m/s). What is the resulting
velocity of the object?
b) Is the collision between the arrow and the object elastic or inelastic? Include
evidence to support your answer. - ANSWERSeq: m1v1 + m2v2 = (m1)(v1A) + (m2)
(V2A)
where: m1 = 0.7 kg, v1 = 90 m/s, m2 = 6.1 kg, v1A = 0.78 m/s
(0.7)(90) + 0 = (0.7)(.78) + (6.1)(v2A)
v2A = 10.24 m/s
b) The initial ad final kinetic energy are not equal (calculated using the following eq:
1/2(m)(v)^2). This means that the collision is inelastic as it lost some kinetic energy
through the collision.
A class F model rocket engine can provide an impulse between 40-80 N-s. A student
attaches a class F rocket that can apply an impulse of 60 N-s to a model rocket of mass
(m = 2 kg) If the engine burns for (t = 7 seconds), what is the average engine thrust
provided by the engine?
b) What maximum speed can the rocket reach? (Assume that air resistance is
negligible.) - ANSWERSeq: J (impulse) = (F)(t)
where: impulse = 60 N-s, t = 7 seconds
(60) = F(7)
,F = 8.57 N
b) eq: J (impulse) = m(delta v)
where: impulse = 60 N-s, m = 2 kg
(60) = (2)(v)
v = 30 m/s
Three particles are held at rest at position (0,0) m. When released, the particles apply a
momentary repelling force on each other. Particle A has a mass of (mA = 1.2 kg) and
has a final velocity of (vA = 2.0 m/s) at 70*. Particle B has a mass of (mB = 1.6 kg)and
has a final velocity of (vB = 3.1 m/s) at 270*. Particle C has a mass of (mC = 2.6 kg).
What are the horizontal and vertical components of particle C's final velocity?
b) What is the velocity of particle C (magnitude and direction)? - ANSWERShorizontal
eq: mv + 0 = (m + m)vx
where: m = 1.2 kg, v = 2 m/s, m2 = 1.6 kg
(1.2)(2) + 0 = (1.2 + 1.6)vx
vx = 0.86 m/s
vertical eq: 0 + mv = (m + m)vy
where: m = 1.6 kg, v = 3.1 m/s, m2 = 1.2 kg
0 + (1.6)(3.1) = (1.6 + 1.2)vy
vy = 1.77 m/s
b) Pythagorean theorem eq: a^2 + b^2 = c^2
where: a = 0.86 m/s, b = 1.77 m/s
(0.86)^2 + (1.77)^2 = v^2
v = 1.97 m/s
tan (1.77/0.86) = 64. 11 degrees
An inattentive car driver crashes into their neighbor's mailbox. Compare the force of the
car on the mailbox to the force of the mailbox on the car. Explain your reasoning.
b) Compare the change in momentum experienced by the car to the change in
momentum experienced by the mailbox. Explain your reasoning.
c) Compare the change in velocity experienced by the car to the change in velocity
experienced by the mailbox. Explain your reasoning. - ANSWERSThe force of the car
, on the mailbox will be equal to the force of the mailbox on the car (when the car hits the
mailbox). This is because of Newton's third law that states applied forces acting on one
another will be equal in magnitude and opposite in direction.
b) The change in momentum of both the car and the mailbox would be the same
according to the conservation of momentum. This is because there are no outside
forces acting on the system that is colliding (the car and the mailbox). This is why we
use the equation "mv = mv".
c) The change in velocity of the mailbox would need to be larger than the change in
velocity of the car. This is because for the systems to be equal, they (obviously) must
equal out. So, since the car has a much larger mass than the mailbox, the mailbox must
have a larger change in velocity.
LESSON 9 QUIZ - ANSWERSstarting now
What maximum horizontal force can a student apply to the top of a 2.0 m cubic box at
rest on a rough surface before the box will start to tip over? The mass of the box is 100
kg. - ANSWERS490 N
A wood board with a length of 2.5 m is allowed to pivot at its midpoint. A 60 kg student
sits 0.8 m from one end of the board. What is the torque of the student on the board? -
ANSWERS265 m-N
A 60 kg student on the end of a 1.5 m diviner's board. What is the student's torque on
the diving board? - ANSWERS882 m-N
Which of the following is NOT an example of an applied net torque? - ANSWERSA
student pushes a box causing it to slide along the ground.
What mass m2 will allow the following mobile to maintain static equilibrium? -
ANSWERS0.075 kg
A bug splats against the windshield of a car traveling at high speeds down a
backcountry road. What statement correctly compares the objects' changes in
momentum? - ANSWERSThe bug's change in momentum is equal to the car's change
in momentum
A baseball bat collides with a 0.145 kg baseball moving with an initial velocity of 35 m/s
to the left. If the collision lasts 0.005 seconds and the final velocity of the ball is 25 m/s
to the right, what was the force of the bat on the ball? - ANSWERS1740 N
A 0.15 kg arrow moving with an initial velocity of 30 m/s passes through a 3.0 kg block
initially at rest on a frictionless surface. If the final speed of the arrow is 25 m/s, what is
the speed of the block? - ANSWERS0.25 m/s
AND ANSWERS 2026
A toy car with mass m1 (0.36 kg ) travels to the right on a frictionless track with a speed
of 3 m/s. A second toy car with a mass m2 (0.74 kg) travels with a speed of 6 m/s to the
left on the same track. The two cars collide and stick together. What are the cars'
speeds after the collision? - ANSWERSmomentum conservation eq:
m1v1 - m2v2 = (m1 + m2)v
where: m1 = 0.36 kg, v1 = 3 m/s, m2 = 0.74 kig, v2 = 6 m/s
(0.36)(3) - (0.74)(6) = (0.36 + 0.74)v
v = 3.05 m/s
A student fires a 0.07 kg arrow at an object with mass m (6.1 kg) that is initially at rest
on a frictionless surface. The speed of the arrow before the collision is 90 m/s. The
speed when the arrow emerges from the object is v (0.78 m/s). What is the resulting
velocity of the object?
b) Is the collision between the arrow and the object elastic or inelastic? Include
evidence to support your answer. - ANSWERSeq: m1v1 + m2v2 = (m1)(v1A) + (m2)
(V2A)
where: m1 = 0.7 kg, v1 = 90 m/s, m2 = 6.1 kg, v1A = 0.78 m/s
(0.7)(90) + 0 = (0.7)(.78) + (6.1)(v2A)
v2A = 10.24 m/s
b) The initial ad final kinetic energy are not equal (calculated using the following eq:
1/2(m)(v)^2). This means that the collision is inelastic as it lost some kinetic energy
through the collision.
A class F model rocket engine can provide an impulse between 40-80 N-s. A student
attaches a class F rocket that can apply an impulse of 60 N-s to a model rocket of mass
(m = 2 kg) If the engine burns for (t = 7 seconds), what is the average engine thrust
provided by the engine?
b) What maximum speed can the rocket reach? (Assume that air resistance is
negligible.) - ANSWERSeq: J (impulse) = (F)(t)
where: impulse = 60 N-s, t = 7 seconds
(60) = F(7)
,F = 8.57 N
b) eq: J (impulse) = m(delta v)
where: impulse = 60 N-s, m = 2 kg
(60) = (2)(v)
v = 30 m/s
Three particles are held at rest at position (0,0) m. When released, the particles apply a
momentary repelling force on each other. Particle A has a mass of (mA = 1.2 kg) and
has a final velocity of (vA = 2.0 m/s) at 70*. Particle B has a mass of (mB = 1.6 kg)and
has a final velocity of (vB = 3.1 m/s) at 270*. Particle C has a mass of (mC = 2.6 kg).
What are the horizontal and vertical components of particle C's final velocity?
b) What is the velocity of particle C (magnitude and direction)? - ANSWERShorizontal
eq: mv + 0 = (m + m)vx
where: m = 1.2 kg, v = 2 m/s, m2 = 1.6 kg
(1.2)(2) + 0 = (1.2 + 1.6)vx
vx = 0.86 m/s
vertical eq: 0 + mv = (m + m)vy
where: m = 1.6 kg, v = 3.1 m/s, m2 = 1.2 kg
0 + (1.6)(3.1) = (1.6 + 1.2)vy
vy = 1.77 m/s
b) Pythagorean theorem eq: a^2 + b^2 = c^2
where: a = 0.86 m/s, b = 1.77 m/s
(0.86)^2 + (1.77)^2 = v^2
v = 1.97 m/s
tan (1.77/0.86) = 64. 11 degrees
An inattentive car driver crashes into their neighbor's mailbox. Compare the force of the
car on the mailbox to the force of the mailbox on the car. Explain your reasoning.
b) Compare the change in momentum experienced by the car to the change in
momentum experienced by the mailbox. Explain your reasoning.
c) Compare the change in velocity experienced by the car to the change in velocity
experienced by the mailbox. Explain your reasoning. - ANSWERSThe force of the car
, on the mailbox will be equal to the force of the mailbox on the car (when the car hits the
mailbox). This is because of Newton's third law that states applied forces acting on one
another will be equal in magnitude and opposite in direction.
b) The change in momentum of both the car and the mailbox would be the same
according to the conservation of momentum. This is because there are no outside
forces acting on the system that is colliding (the car and the mailbox). This is why we
use the equation "mv = mv".
c) The change in velocity of the mailbox would need to be larger than the change in
velocity of the car. This is because for the systems to be equal, they (obviously) must
equal out. So, since the car has a much larger mass than the mailbox, the mailbox must
have a larger change in velocity.
LESSON 9 QUIZ - ANSWERSstarting now
What maximum horizontal force can a student apply to the top of a 2.0 m cubic box at
rest on a rough surface before the box will start to tip over? The mass of the box is 100
kg. - ANSWERS490 N
A wood board with a length of 2.5 m is allowed to pivot at its midpoint. A 60 kg student
sits 0.8 m from one end of the board. What is the torque of the student on the board? -
ANSWERS265 m-N
A 60 kg student on the end of a 1.5 m diviner's board. What is the student's torque on
the diving board? - ANSWERS882 m-N
Which of the following is NOT an example of an applied net torque? - ANSWERSA
student pushes a box causing it to slide along the ground.
What mass m2 will allow the following mobile to maintain static equilibrium? -
ANSWERS0.075 kg
A bug splats against the windshield of a car traveling at high speeds down a
backcountry road. What statement correctly compares the objects' changes in
momentum? - ANSWERSThe bug's change in momentum is equal to the car's change
in momentum
A baseball bat collides with a 0.145 kg baseball moving with an initial velocity of 35 m/s
to the left. If the collision lasts 0.005 seconds and the final velocity of the ball is 25 m/s
to the right, what was the force of the bat on the ball? - ANSWERS1740 N
A 0.15 kg arrow moving with an initial velocity of 30 m/s passes through a 3.0 kg block
initially at rest on a frictionless surface. If the final speed of the arrow is 25 m/s, what is
the speed of the block? - ANSWERS0.25 m/s
