A bullet of mass ml >.050kg strikes and passes through a wooden block whose mass

Question

A bullet of mass ml >.050kg strikes and passes through a wooden block whose mass is m2 = 4.0kg. The block is suspended vertically by a 1 m long string. The bullet emerges with a speed of 60 m/s, and the block swings through a maximum angle of 32.46 degrees. a) Is kinetic energy conserved in the impact>? b) Is momentum conserved in the impact? c) How high is the block at the highest point? C) What is the velocity of the block just after the bullet passed through it? d) What was the initial velocity of the bullet.

Explanation / Answer

The maximum angle is 32.46 degree then the maximum height will be L(1-cos(theta)) = 1*(1-cos(32-46 deg)) Maximum height = 0.156 m

So the work done by gravity = -mgh = -4*9.81*0.156 = -6.12 J

So the work energy theorem will be :

workdone = change in kinetic energy

-6.12 = 0-0.5*4*v^2 = -2v^2

velocity of the block just after the collision v = 1.75 m/s

In each collision the momentum remains conserved, So applying momentum conservation,

0+0.05*v' = 4*1.75+0.05*60

Initial velocity of the bullet v'= 200 m/s to the right.

So restitution constant e = (60+1.75)/200 =0.30875

Since e<1 hence it is not purily elastic hence kinetic energy won't be conserved.

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