A 12 g bullet passes horizontally through a 15 cm tall, 20 cm long .5 kg block w

Question

A 12 g bullet passes horizontally through a 15 cm tall, 20 cm long .5 kg block which is initially at rest on a 1 meter tall table's edge. The bullet is initially moving 230 m/s, and it passes through the block. After the bullet passes through it, the block hits the ground 134 cm from the edge of the table.

How fast is the block moving after the bullet emerges from it?

How fast is the bullet moving when it merges from the block?

Assuming that the bolck does not move significantly while the bullet is inside it, determine the average force exerted on the block by the bullet while the bullet passes through the block. Use energy!

Explanation / Answer

Given that Horizontal range of the block is given by R = u*sqrt(2h/g) = 134 cm = 1.34

u*sqrt(2*h/g) = 1.34

u*sqrt(2*1/9.8) = 1.34

u = 2.96 m/sec is the speed of the block after the bullet emerges out

Using law of conservation of momentum

(m*Vo) = m*V + (M*u)

(12*10^-3*230) = (12*10^-3*V) + (0.5*2.96)

V = 106.67 m/sec is the speed of the bullet when it emerges from the block

Using law of conservation of energy

Work done by the block = chnage in Kinetic energy of the block

Favg*S = 0.5*m*(V^2-Vo^2)

Favg*0.2 = 0.5*0.5*(106.67^2 - 230^2) = -10380.37

Favg = -10380.37/0.2 = -51901.85 N (action)

So by the Newton's third law of motion

action = -reaction

then Force exerted by the block on bullet is 51901.85 N

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