A block of wood (M block = 5.3 kg) rests on a horizontal surface for which the c

Question

A block of wood (Mblock = 5.3 kg) rests on a horizontal surface for which the coefficient of kinetic friction is 0.21. A bullet (mbullet = 37 g) is shot along the horizontal direction and embeds in the block, displacing the block 86 cm. What is the magnitude of the acceleration of the bullet and block from immediately after the impact until they come to a stop? What is the magnitude of the velocity of the bullet and block immediately after the impact? What was the magnitude of the initial velocity of the bullet immediately before the impact?

Explanation / Answer

the only force acting on the block and bullet is frictional force

fk = -uk*( M + m)*g

acceleration = a = fk/(M+m) = -uk*g = 0.21*9.8 = -2.058 m/s^2

magnitude = 2.058 m/s^2 <<_----answer

(b)

final velocity = v2 = 0

displacement = x = 0.86 m

from equations of motion

v2^2 - v1^2 = 2*a*x

0 - v1^2 = -2*2.058*0.86

v1 = 1.88 m/s <-------answer

(b)

from conservation of momentum

momentum before collision = momentum after collision

M*V + m*U = (M+m)*V1

V = 0

0.037*u = (5.3+0.037)*1.88

u = 271.1 m/s <<-answer

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