A 5.00-g bullet moving with an initial speed of vi = 390 m/s is fired into and p

Q: A 5.00-g bullet moving with an initial speed of vi = 390 m/s is fired into and p

kinetic energy of the block = elastic potential energy 0.5*M*v^2 = 0.5*K*d^2 0.5*1*V^2 = 0.5*890*0.054^2 V = 1.61 m/s from conservation of momentum momentum before collision = momentum after collision m*vi = M*V + m*vf 0.005*390 = (1*1.61) + (0.005*vf) vf = 68 m/s

ID: 1512152 • Letter: A

Question

A 5.00-g bullet moving with an initial speed of vi = 390 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 890 N/m. The block moves d = 5.40 cm to the right after impact before being brought to rest by the spring. (a) Find the speed at which the bullet emerges from the block. m/s (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision.

Explanation / Answer

kinetic energy of the block = elastic potential energy

0.5*M*v^2 = 0.5*K*d^2

0.5*1*V^2 = 0.5*890*0.054^2

V = 1.61 m/s

from conservation of momentum

momentum before collision = momentum after collision

m*vi = M*V + m*vf

0.005*390 = (1*1.61) + (0.005*vf)

vf = 68 m/s <<<-----answer

internal energy = KEi - KEf

internal energy = 0.5*0.005*(390^2-68^2) = 368.69 J