A 0.02kg bullet traveling at 200m/s penetrates a 2kg block of wood and dremains

Question

A 0.02kg bullet traveling at 200m/s penetrates a 2kg block of wood and dremains embedded in the block. If the block is stationary on a frictionless surface when hit, answer:

a) Write the correct equation that describes the conservation of the total momentum for the sytem (bullet + block)

b) Use the equation at point a) to calculate the velocity of the block + bullet ensemble after the collision

c) What is the impulse delivered to the block by the bullet?

d) If the collision time (time it takes the bullet to enter the block) is 0.01s, what is the average force exerted on the block?

Explanation / Answer

A) m1u1 + m2u2 = mfvf

but, since the initial velocity of the block is 0, it can be further simplified to

m1u1 = mfvf

B) 0.02 x 200 = (0.02 + 2) x vf

vf = 4/2.02

= 1.98 m/s

C) I = mv

= 0.02 x (200 - 1.98)

= 3.9604 kgm/s

D) a = v-u/t

a = 198.02 / 0.01

= 19802 m/s/s

F = ma

= 0.02 x 19802

= 296.04 N

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