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Energy & momentum: Physics Regents Practice

Practice Energy & momentum for the Physics Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 33 · 2 pts

A $1500$-kilogram car travels along a straight road at $20$ meters per second. What is the magnitude of the car's momentum?

Worked solution
$p = mv$ Momentum relationship from the reference tables.
$p = (1500\ \text{kg})(20\ \text{m/s})$
$p = 3.0 \times 10^{4}\ \text{kg}\cdot\text{m/s}$
Answer: $3.0 \times 10^{4}$ kg·m/s

Momentum is the product of mass and velocity, $p = mv$, a vector pointing in the direction of motion.

Question 34 · 2 pts

A $2.0$-kilogram ball moves at $3.0$ meters per second. What is its kinetic energy?

Worked solution
$KE = \frac{1}{2}mv^2$ Kinetic energy from the reference tables.
$KE = \frac{1}{2}(2.0)(3.0)^2$
$KE = \frac{1}{2}(2.0)(9.0) = 9.0\ \text{J}$
Answer: $9.0$ J

Kinetic energy grows with the square of speed, $KE = \frac{1}{2}mv^2$, so doubling speed quadruples the energy.

Question 48 · 2 pts

A $0.50$ kg ball is released from rest at the top of a frictionless ramp $1.8$ m above the ground. What is the ball's speed at the bottom? ($g = 9.81$ m/s$^2$)

Worked solution
$mgh = \tfrac{1}{2}mv^2$ Energy conservation; the mass cancels from both sides.
$v = \sqrt{2gh} = \sqrt{2(9.81)(1.8)}$
$v = \sqrt{35.3} = 5.9\ \text{m/s}$
Answer: $5.9$ m/s

Gravitational potential energy converts fully to kinetic energy on a frictionless ramp; the mass cancels, so the bottom speed depends only on the drop height.

Question 5 · 2 pts

What is the kinetic energy of a $2.0\ \text{kg}$ object moving at $3.0\ \text{m/s}$?

Worked solution
$KE = \tfrac{1}{2}mv^2$ Kinetic energy formula
$KE = \tfrac{1}{2}(2.0\ \text{kg})(3.0\ \text{m/s})^2$
$KE = \tfrac{1}{2}(2.0)(9.0) = 9.0\ \text{J}$
Answer: $9.0\ \text{J}$

Kinetic energy is one-half the mass times the square of the speed; squaring 3.0 m/s first and then applying the one-half factor gives 9.0 J.

Question 44 · 2 pts

A 2.0 kg cart moving at 6.0 m/s collides with a stationary 4.0 kg cart, and the two carts stick together after the collision. What is their common speed immediately after the collision?

Worked solution
$\text{Conservation of momentum: } p_i = p_f$
$m_A v_A + m_B v_B = (m_A + m_B) v_f$
$(2.0)(6.0) + (4.0)(0) = (2.0 + 4.0) v_f$
$12 = 6.0\, v_f \Rightarrow v_f = 2.0\ \text{m/s}$
Answer: 2.0 m/s

In a perfectly inelastic collision, total momentum before equals total momentum after, with the combined mass moving at one common velocity; dividing the initial momentum (12 kg·m/s) by the total mass (6.0 kg) gives 2.0 m/s. Using only one cart's mass in the denominator produces the wrong distractors.

Question 18 · 2 pts

A 2.0-kilogram object falls freely from rest through a height of 5.0 m. What is its kinetic energy just before it strikes the ground? (Neglect air resistance.)

Worked solution
$\text{By conservation of energy, } KE_f = PE_i = mgh$
$KE_f = (2.0\ \text{kg})(9.81\ \text{m/s}^2)(5.0\ \text{m})$
$KE_f = 98.1\ \text{J} \approx 98\ \text{J}$
Answer: 98 J

All of the gravitational potential energy converts to kinetic energy as the object falls, so $KE = mgh$. Multiplying by an extra factor of 2 (196 J) is a frequent arithmetic slip.

Question 30 · 2 pts

A 2.0-kg cart moving at $3.0\ \text{m/s}$ collides with a stationary 4.0-kg cart, and the two carts stick together after the collision. What is the speed of the combined carts immediately after the collision?

Worked solution
$p_{before} = m_1 v_1 + m_2 v_2 = (2.0)(3.0) + (4.0)(0) = 6.0\ \text{kg}\cdot\text{m/s}$
$p_{before} = p_{after} = (m_1+m_2)v'$ Conservation of momentum, perfectly inelastic collision
$v' = \dfrac{6.0}{2.0+4.0} = 1.0\ \text{m/s}$
Answer: $1.0\ \text{m/s}$

In a perfectly inelastic collision, the two objects move together afterward, so total momentum before equals the combined mass times the common final velocity.

Question 41 · 2 pts

What is the kinetic energy of a 2.0 kg object moving at a speed of 3.0 m/s?

Worked solution
$KE = \tfrac{1}{2}mv^2$
$KE = \tfrac{1}{2}(2.0\text{ kg})(3.0\text{ m/s})^2$
$KE = \tfrac{1}{2}(2.0)(9.0) = 9.0\text{ J}$
Answer: 9.0 J

Kinetic energy equals one-half the mass times the square of the speed; squaring the speed before multiplying gives 9.0 J, whereas forgetting to square the speed or forgetting the factor of one-half leads to the incorrect distractors.

Question 6 · 2 pts

A $2.0\ \text{kg}$ object is raised to a height of $10.\ \text{m}$ above the ground. What is its gravitational potential energy relative to the ground?

Worked solution
$PE = mgh$
$PE = (2.0\ \text{kg})(9.81\ \text{m/s}^2)(10.\ \text{m})$
$PE = 196\ \text{J}$
Answer: $196\ \text{J}$

Gravitational potential energy equals mass times gravitational acceleration times height; multiplying all three given values yields 196 J.

Question 33 · 2 pts

A 2.0 kg object is dropped from rest and falls freely for $5.0\ \text{m}$. Using conservation of energy, what is the object's speed just before it strikes the ground?

Worked solution
$PE_{lost} = KE_{gained}$ conservation of mechanical energy
$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$
$v = \sqrt{2(9.81\ \text{m/s}^2)(5.0\ \text{m})} = \sqrt{98.1}\ \text{m/s}$
$v = 9.9\ \text{m/s}$
Answer: 9.9 m/s

All the gravitational potential energy converts to kinetic energy, giving $v=\sqrt{2gh}$. Forgetting to take the square root leads to the distractor 98 m/s, and using half the height gives 4.9 m/s.

Question 45 · 2 pts

What is the kinetic energy of a 2.0 kg object moving with a speed of 6.0 m/s?

Worked solution
$KE = \tfrac{1}{2}mv^2$ Kinetic energy formula
$KE = \tfrac{1}{2}(2.0\ \text{kg})(6.0\ \text{m/s})^2 = \tfrac{1}{2}(2.0)(36)$
$KE = 36\ \text{J}$
Answer: 36 J

Kinetic energy is one-half the mass times the square of the speed. Squaring 6.0 m/s gives 36, and multiplying by half of 2.0 kg gives 36 J. Forgetting the factor of one-half yields the distractor 72 J.

Question 55 · 4 pts · constructed response

A 0.500 kg ball is dropped from rest from a height of 10.0 m above the ground. Using conservation of energy, calculate the speed of the ball just before it strikes the ground, then calculate the time it takes to fall.

Model answer: The ball's speed just before impact is $14.0\ \text{m/s}$ and the fall takes $1.43\ \text{s}$.
Worked solution
$mgh = \tfrac{1}{2}mv^2$ All gravitational PE converts to KE (no air resistance)
$v = \sqrt{2gh} = \sqrt{2(9.81)(10.0)} = \sqrt{196.2}$
$v = 14.0\ \text{m/s}$
$v = gt \Rightarrow t = \dfrac{v}{g} = \dfrac{14.0}{9.81}$ Ball starts from rest, so v = gt
$t = 1.43\ \text{s}$
Answer: The ball's speed just before impact is $14.0\ \text{m/s}$ and the fall takes $1.43\ \text{s}$.

Since the ball starts at rest and only gravity acts on it, all of its initial gravitational potential energy converts to kinetic energy at ground level, giving its final speed, and that speed together with v = gt gives the fall time.

Question 3 · 2 pts

A 2.0 kg object moves with a speed of 6.0 m/s. What is its kinetic energy?

Worked solution
$KE = \tfrac{1}{2}mv^2$ kinetic energy formula
$KE = \tfrac{1}{2}(2.0)(6.0)^2 = \tfrac{1}{2}(2.0)(36)$
$KE = 36\ J$
Answer: 36 J

Kinetic energy depends on the square of speed; substituting the given mass and speed yields 36 J.

Question 44 · 2 pts

What is the kinetic energy of a 2.0-kilogram object moving at a speed of 4.0 meters per second?

Worked solution
$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2.0)(4.0)^2 = 16\ \text{J}$ Kinetic energy formula
Answer: 16 J

Kinetic energy is one-half the mass times the square of the speed, giving 16 J.

Question 20 · 1 pt

What is the kinetic energy of a $2\ \mathrm{kg}$ object moving at $3\ \mathrm{m/s}$ ? (Use $KE = \tfrac{1}{2}mv^2$.)

Worked solution
$KE = \tfrac12 m v^2 = \tfrac12 (2)(3)^2$ Substitute m = 2, v = 3.
$= \tfrac12 (2)(9) = 9\ \mathrm{J}$
Answer: 9\ \mathrm{J}

Kinetic energy depends on the square of speed: KE = ½mv². Square the velocity before multiplying.

Question 35 · 2 pts

A constant $50$-newton force acts for $4.0$ seconds on a $10$-kilogram object that starts from rest on a frictionless surface. What is the object's final speed?

Worked solution
$J = Ft = \Delta p = m\Delta v$ Impulse equals the change in momentum.
$Ft = (50)(4.0) = 200\ \text{N}\cdot\text{s}$
$v = \frac{Ft}{m} = \frac{200}{10} = 20\ \text{m/s}$
Answer: $20$ m/s

Impulse, the product of force and time, equals the change in momentum, $Ft = m\Delta v$.

Question 36 · 2 pts

A $2.0$-kilogram cart moving at $4.0$ meters per second collides with and sticks to a $2.0$-kilogram cart at rest. What is the speed of the combined carts immediately after the collision?

Worked solution
$p_{before} = p_{after}$ Momentum is conserved in the collision.
$m_1 v_1 = (m_1 + m_2)v_f$
$(2.0)(4.0) = (4.0)\,v_f$
$v_f = \frac{8.0}{4.0} = 2.0\ \text{m/s}$
Answer: $2.0$ m/s

In a perfectly inelastic collision the objects move together, and total momentum before equals total momentum after.

Question 38 · 2 pts

A $0.50$-kilogram ball is dropped from rest at a height of $8.0$ meters. Neglecting air resistance, what is its speed just before it strikes the ground?

Worked solution
$mgh = \frac{1}{2}mv^2$ Gravitational PE converts to kinetic energy; mass cancels.
$v = \sqrt{2gh}$
$v = \sqrt{2(9.81)(8.0)} = \sqrt{156.96}$
$v \approx 12.5\ \text{m/s}$
Answer: $12.5$ m/s

Energy conservation gives $v = \sqrt{2gh}$, a result independent of mass because mass cancels out.

Question 41 · 2 pts

A $0.20$ kg ball moving to the right at $15$ m/s strikes a wall and rebounds straight back to the left at $10$ m/s. What is the magnitude of the change in the ball's momentum during the collision?

Worked solution
$p = mv, \qquad J = \Delta p$ Impulse equals change in momentum (NYS reference table).
$v_i = +15\ \text{m/s}, \qquad v_f = -10\ \text{m/s}$ A rebound reverses direction, so the velocities have opposite signs.
$\Delta p = m(v_f - v_i) = 0.20(-10 - 15)$
$\Delta p = 0.20(-25) = -5.0\ \text{kg·m/s}$
$|\Delta p| = 5.0\ \text{kg·m/s}$
Answer: $5.0$ kg·m/s

Because the ball reverses direction, the change in momentum uses the difference of oppositely signed velocities, making it larger than either momentum alone.

Question 42 · 2 pts

A $4.0$ kg rifle, initially at rest, fires a $0.020$ kg bullet horizontally at $300$ m/s. What is the recoil speed of the rifle?

Worked solution
$p_{before} = p_{after} = 0$ The rifle and bullet start at rest, so total momentum is zero.
$0 = m_b v_b + m_r v_r$
$v_r = -\frac{m_b v_b}{m_r} = -\frac{(0.020)(300)}{4.0}$
$v_r = -\frac{6.0}{4.0} = -1.5\ \text{m/s}$
Answer: $1.5$ m/s, directed opposite to the bullet

Total momentum is conserved and starts at zero, so the rifle's momentum is equal and opposite to the bullet's. Because the rifle is far more massive, its recoil speed is small.

Question 43 · 2 pts

A spring of force constant $k = 200$ N/m is compressed $0.10$ m and used to launch a $0.50$ kg cart along a frictionless horizontal track. What speed does the cart reach?

Worked solution
$PE_s = \tfrac{1}{2}kx^2$ Elastic potential energy from the reference table.
$PE_s = \tfrac{1}{2}(200)(0.10)^2 = 1.0\ \text{J}$
$PE_s = KE = \tfrac{1}{2}mv^2$ On a frictionless track all spring PE becomes KE.
$v = \sqrt{\frac{2\,PE_s}{m}} = \sqrt{\frac{2(1.0)}{0.50}}$
$v = \sqrt{4.0} = 2.0\ \text{m/s}$
Answer: $2.0$ m/s

Energy stored in the compressed spring converts entirely into kinetic energy of the cart on the frictionless surface.

Question 45 · 2 pts

A motor lifts a $50$ kg load straight up at a constant speed of $2.0$ m/s. What is the power output of the motor? ($g = 9.81$ m/s$^2$)

Worked solution
$P = Fv$ Power as force times velocity (reference table).
$F = mg = (50)(9.81) = 490.5\ \text{N}$ Constant speed means the lift force equals the weight.
$P = (490.5)(2.0)$
$P = 981\ \text{W} \approx 9.8\times10^2\ \text{W}$
Answer: $981$ W (about $9.8\times10^2$ W)

At constant velocity the applied force equals the weight, so the power delivered is weight multiplied by lifting speed.

Question 47 · 2 pts

A $2.0$ kg cart moving at $3.0$ m/s collides with and sticks to a stationary $1.0$ kg cart on a frictionless track. How much kinetic energy is lost in the collision?

Worked solution
$m_1 v_1 = (m_1 + m_2)v_f$ Momentum is conserved in the collision.
$v_f = \frac{(2.0)(3.0)}{3.0} = 2.0\ \text{m/s}$
$KE_i = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$
$KE_f = \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}$
$\Delta KE = 9.0 - 6.0 = 3.0\ \text{J}$
Answer: $3.0$ J

Momentum is conserved in every collision, but in a perfectly inelastic (sticking) collision some kinetic energy is converted to heat and deformation.

Question 6 · 2 pts

A $2.0\ \text{kg}$ cart moving at $4.0\ \text{m/s}$ collides with and sticks to an identical $2.0\ \text{kg}$ cart initially at rest. What is the speed of the combined carts immediately after the collision?

Worked solution
$p_i = p_f$ Conservation of momentum for a perfectly inelastic collision
$m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f$
$(2.0)(4.0) + (2.0)(0) = (4.0)v_f$
$8.0 = 4.0\,v_f \Rightarrow v_f = 2.0\ \text{m/s}$
Answer: $2.0\ \text{m/s}$

Total momentum before the collision equals total momentum after; since the carts stick together, dividing the initial momentum by the combined mass gives the common final speed of 2.0 m/s.

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