The ballistic pendulum consists of a stationary 2.5 kg block of wood suspended b

Question

The ballistic pendulum consists of a stationary 2.5 kg block of wood suspended by a wire of negligible mass. A 0.01 kg bullet is fired into the block, and then the block, which was at rest before the collision, swings to a maximum height of 0.65 m above the initial position. Using the principle of momentum conservation, find the expression for the speed of the bullet right before it collides (v_01) with the wooden block. Use v_f for the speed of the block after the collision. Using the principle of conservation of mechanical energy, find the value of v_f. What is the value of v_01?

Explanation / Answer

a)

m1 = mass of bullet = 0.01 kg

m2 = mass of block = 2.5 kg

Vo1 = velocity of bullet before collision

Vo2 = velocity of block before collision = 0

Vf = combined velocity of bullet + block after collision

Using conservation of momentum

m1 Vo1 + m2 Vo2 = (m1 + m2) Vf

Vf = m1 Vo1 /(m1 + m2)

b)

Using conservation of energy :

Kinetic energy after collision = potential energy at the top of height "h"

(0.5) m Vf2 = m g h

Vf2 = 2 gh

Vf2 = 2 x 9.8 x 0.65

Vf = 3.57 m/s

c)

Vf = m1 Vo1 /(m1 + m2)

3.57 = 0.01 Vo1 / (0.01 + 2.5)

Vo1 = 896.1 m/s

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