Zaid Berjis

PhysicsUnit review 18 of 18About 50 minutes

Review: momentum and collisions

This review pulls together momentum, impulse and collisions. The first set of problems comes with a simulation: solve the problem on paper first, then run the simulation to check your answer.

Key ideas of the unit

IdeaEquationLesson
Momentum\(\vec p = m\vec v\)15
Impulse-momentum theorem\(\vec F\Delta t = \Delta\vec p\)15
Conservation of momentum\(m_1\vec v_{1,i} + m_2\vec v_{2,i} = m_1\vec v_{1,f} + m_2\vec v_{2,f}\)16
Perfectly inelastic collision\(m_1\vec v_{1,i} + m_2\vec v_{2,i} = (m_1 + m_2)\vec v_f\); KE lost17
Elastic collisionMomentum and KE both conserved17

Problems with simulations

Problem 1 · Recoil

A 40.0 kg tennis-ball launcher sits at rest on a frictionless surface. A spring inside it fires a 58.0 g ball north at 30.0 m/s. What is the launcher's recoil velocity?

  1. Total momentum before: zero, because everything is at rest.
  2. Conservation, north positive. \(0 = (0.0580)(30.0) + (40.0)v\)
  3. Solve. \(v = -\dfrac{1.74}{40.0} = -0.0435\ \text{m/s}\)

    0.0435 m/s south

Ball launcher recoil

Simulation

Problem 2 · Collision, then projectile motion

A 6.0 g dart moving horizontally at 20.0 m/s sticks in a 24.0 g block resting on top of a 1.20 m post. The block and dart fly off together. How far from the base of the post do they land?

  1. The collision is perfectly inelastic. \(v_f = \dfrac{(0.0060)(20.0)}{0.0060 + 0.0240} = \dfrac{0.120}{0.0300} = 4.00\ \text{m/s}\), horizontal.
  2. Time to fall 1.20 m from rest vertically: \(t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2(1.20)}{9.81}} = 0.495\ \text{s}\).
  3. Horizontal distance. \(\Delta x = v_f t = (4.00)(0.495) = 1.98\ \text{m}\)

    About 1.98 m from the post

Dart into a block on a post

Simulation

Try this

Fire at 20 m/s to check your answer, then predict the landing distance for 30 m/s before you try it.

Problem 3 · Momentum, then energy

A 40.0 g bird sits on a 120 g swing seat that hangs at rest. The bird takes off horizontally at 2.40 m/s. How high does the swing seat rise?

  1. Push-off conserves momentum. \(0 = (0.0400)(2.40) + (0.120)v\), so the seat moves backward at \(v = 0.800\ \text{m/s}\).
  2. Then the swing conserves mechanical energy. \(\tfrac12 mv^2 = mgh\), so \(h = \dfrac{v^2}{2g} = \dfrac{(0.800)^2}{2(9.81)} = 0.0326\ \text{m}\).

    About 3.3 cm. (The simulation shows 3.2 cm, because its push lasts a few milliseconds and the seat starts to swing during it.)

A bird pushes off a swing

Simulation

Problem 4 · Ballistic pendulum

A 10.0 g bullet moving at 300 m/s embeds itself in a 2.00 kg wooden block hanging from a 1.0 m string. (a) How fast do the block and bullet move just after the impact? (b) How high do they swing? (c) What fraction of the bullet's kinetic energy is lost?

  1. (a) Perfectly inelastic collision. \(V = \dfrac{(0.0100)(300)}{2.010} = 1.49\ \text{m/s}\)
  2. (b) Then energy is conserved in the swing. \(h = \dfrac{V^2}{2g} = \dfrac{(1.49)^2}{2(9.81)} = 0.114\ \text{m}\)
  3. (c) Kinetic energy. Before: \(\tfrac12(0.0100)(300)^2 = 450\ \text{J}\). After: \(\tfrac12(2.010)(1.49)^2 = 2.24\ \text{J}\). Lost: \(\dfrac{450 - 2.24}{450} = 99.5\%\).

    (a) 1.49 m/s. (b) 11.4 cm. (c) 99.5% of the kinetic energy becomes heat and deformation.

Ballistic pendulum

Simulation

Problem 5 · Elastic collision

A 0.300 kg bead slides along a frictionless wire at 4.00 cm/s and collides elastically with a 0.500 kg bead at rest. Afterward the small bead moves at 1.00 cm/s in the opposite direction. Find the velocity of the large bead and confirm that kinetic energy is conserved.

  1. Momentum (in kg·cm/s). \((0.300)(4.00) = (0.300)(-1.00) + (0.500)v_2\), so \(1.20 = -0.300 + 0.500v_2\) and \(v_2 = 3.00\ \text{cm/s}\).
  2. Kinetic energy (convert to m/s). Before: \(\tfrac12(0.300)(0.0400)^2 = 2.40 \times 10^{-4}\ \text{J}\). After: \(\tfrac12(0.300)(0.0100)^2 + \tfrac12(0.500)(0.0300)^2 = 0.15 \times 10^{-4} + 2.25 \times 10^{-4} = 2.40 \times 10^{-4}\ \text{J}\).

    3.00 cm/s forward; kinetic energy is the same before and after.

Beads on a wire

Simulation

Problem 6 · Momentum in two dimensions

A nucleus of mass \(20.0 \times 10^{-27}\ \text{kg}\), at rest, breaks into three pieces. Piece 1 (\(6.0 \times 10^{-27}\ \text{kg}\)) moves along \(+y\) at \(5.0 \times 10^6\ \text{m/s}\). Piece 2 (\(9.0 \times 10^{-27}\ \text{kg}\)) moves along \(+x\) at \(3.0 \times 10^6\ \text{m/s}\). Find the speed and direction of piece 3.

  1. Total momentum stays zero in both directions, so piece 3 must cancel the other two.
  2. Momenta of pieces 1 and 2. \(p_{1y} = (6.0 \times 10^{-27})(5.0 \times 10^6) = 3.0 \times 10^{-20}\) and \(p_{2x} = (9.0 \times 10^{-27})(3.0 \times 10^6) = 2.7 \times 10^{-20}\) (kg·m/s).
  3. Piece 3. \(p_{3x} = -2.7 \times 10^{-20}\), \(p_{3y} = -3.0 \times 10^{-20}\). Size: \(\sqrt{2.7^2 + 3.0^2} \times 10^{-20} = 4.04 \times 10^{-20}\ \text{kg·m/s}\).
  4. Speed and direction. Its mass is \((20.0 - 6.0 - 9.0) \times 10^{-27} = 5.0 \times 10^{-27}\ \text{kg}\), so \(v_3 = \dfrac{4.04 \times 10^{-20}}{5.0 \times 10^{-27}} = 8.1 \times 10^6\ \text{m/s}\), at \(\tan^{-1}(3.0/2.7) = 48^\circ\) below the \(-x\) axis.

    \(8.1 \times 10^6\ \text{m/s}\), 48° below the negative \(x\)-axis

A nucleus splits in three

Simulation

Practice questions

Question 1

A catcher stops a 0.145 kg baseball moving at 40.0 m/s. The ball pushes the glove back for 0.100 s. What average force does the glove exert?

Show solution

\(F = \dfrac{\Delta p}{\Delta t} = \dfrac{(0.145)(40.0)}{0.100} = 58.0\ \text{N}\)

Question 2

A 0.300 kg ball hits the floor at 6.00 m/s and rebounds straight up at 4.00 m/s. What is the size of the impulse the floor gives the ball?

Show solution

Up is positive: \(\Delta p = 0.300(4.00 - (-6.00)) = 3.00\ \text{N·s}\), upward.

Question 3

A 90.0 kg football player running at 5.00 m/s tackles a 75.0 kg player standing still, and they move off together. How fast do they move just after the tackle?

Show solution

\(v_f = \dfrac{(90.0)(5.00)}{165} = 2.73\ \text{m/s}\)

Question 4

A bowling ball and a tennis ball have the same momentum. Which has more kinetic energy?

  1. The bowling ball
  2. They have the same kinetic energy
  3. It depends on the direction of motion
  4. The tennis ball
Show solution

\(KE = \dfrac{p^2}{2m}\). With the same \(p\), the smaller mass has more kinetic energy. The tennis ball must be moving much faster.

Question 5

A gymnast lands on a thick mat instead of a hard floor. Compared with the floor, the mat...

  1. increases the stopping time, so the average force is smaller
  2. reduces the gymnast's change in momentum
  3. reduces the impulse on the gymnast
  4. increases the gymnast's momentum before landing
Show solution

The change in momentum and the impulse are fixed by the landing speed. The mat stretches the time, which lowers the average force.

Where next?

  • You have finished the mechanics course. Go back to the Physics page to check which lessons you have marked complete.
  • For a challenge, revisit any simulation above and change the numbers, then predict the outcome before you run it.

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