Question

OAL Combine the concepts of conservation of energy and conservation of momentum in inelastic collisions.

In figure a, a bullet and a wooden block are shown in two configurations.

In the first configuration, the block, labeled m2, hangs vertically from a ceiling. A bullet, labeled m1, approaches the block horizontally from the left. A rightward arrow points from the bullet and is labeled vector v1i. A rightward arrow, shorter than the first, points from the block and is labeled vector vsys.

In the second configuration, the bullet is embedded in the block, and the block has swung up and to the right, a vertical distance h from its initial position. The block and bullet are labeled m1 + m2.

In figure b, a photograph shows a laboratory ballistic pendulum. A person pulls a trigger on the left, and a pendulum is show in multiple positions as it swings up and to the right.

(a) Diagram of a ballistic pendulum. Note that v with arrowsys is the velocity of the system just after the perfectly inelastic collision. (b) Multiflash photograph of a laboratory ballistic pendulum.

PROBLEM The ballistic pendulum (Figure a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the technique, assume that the mass of the bullet, m1, is 5.00 g, the mass of the pendulum, m2, is 1.000 kg, and h is 5.00 cm. (a) Find the velocity of the system after the bullet embeds in the block. (b) Calculate the initial speed of the bullet.

STRATEGY Use conservation of energy to find the initial velocity of the block–bullet system, labeling it vsys. Part (b) requires the conservation of momentum equation, which can be solved for the initial velocity of the bullet, v1i.

SOLUTION

(A) Find the velocity of the system after the bullet embeds in the block.

Apply conservation of energy to the block–bullet system after the collision.

(KE + PE)after collision = (KE + PE)top

Substitute expressions for the kinetic and potential energies. Note that both the potential energy at the bottom and the kinetic energy at the top are zero.

1

2

(m1 + m2)vsys2 + 0 = 0 + (m1 + m2)gh

Solve for the final velocity of the block–bullet system, vsys.

vsys2 = 2gh

vsys =

2gh

=

2(9.80 m/s2)(5.00 ✕ 10−2 m)

vsys = 0.990 m/s

(B) Calculate the initial speed of the bullet.

Write the conservation of momentum equation and substitute expressions.

pi = pf

m1v1i + m2v2i = (m1 + m2)vsys

Solve for the initial velocity of the bullet, and substitute values.

v1i =

(m1 + m2)vsys

m1

v1i =

(1.005 kg)(0.990 m/s)

5.00 ✕ 10−3 kg

= 199 m/s

LEARN MORE

REMARKS Because the impact is inelastic, it would be incorrect to equate the initial kinetic energy of the incoming bullet to the final gravitational potential energy associated with the bullet–block combination. The energy isn't conserved!

QUESTION The ways that mechanical energy is lost from the system in this experiment include: (Select all that apply.)

friction in the mechanisms

thermal energy loss due to air drag

energy loss from the change in height of the block

friction to bring the bullet to a stop relative to the ground

emission of sound waves

PRACTICE IT

Use the worked example above to help you solve this problem. The ballistic pendulum (Figure a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the technique, assume that the mass of the bullet, m1, is 4.51 g, the mass of the pendulum, m2, is 1.051 kg, and h is 4.84 cm.

(a) Find the velocity of the system after the bullet embeds in the block.

m/s

(b) Calculate the initial speed of the bullet.

m/s

EXERCISE HINTS: GETTING STARTED | I'M STUCK!

A bullet with mass 5.19 g is fired horizontally into a 2.066-kg block attached to a horizontal spring. The spring has a constant 6.09 102 N/m and reaches a maximum compression of 5.99 cm.

(a) Find the initial speed of the bullet–block system.

m/s

(b) Find the speed of the bullet.

m/s

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Answer 1

a)

m₁ = mass of bullet = 4.51 g = 0.00451 kg

v_1i = initial speed of the bullet before collision = ?

m₂ = mass of pendulum = 1.051 kg

v_2i = initial speed of the pendulum before collision = 0 m/s

v = speed of the combination just after collision

h = height = 4.84 cm = 0.0484 m

Using conservation of energy

Gravitational potential energy gained = Kinetic energy lost

(m₁ + m₂) gh = (0.5) (m₁ + m₂) v²

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 0.0484)

v = 0.974 m/s

b)

Using conservation of momentum

m₁ v_1i + m₂ v_2i = (m₁ + m₂) v

(0.00451) v_1i + (1.051) (0) = (0.00451 + 1.051) (0.974)

v_1i = 228 m/s