Posted by Anonymous on .
A 7 x 10^-3kg bullet is fired into a 2kg wooden block initially at rest on a level surface. The bullet passes through the block and emerges with the velocity of 200m/s. The block slides 0.5m and stops. If the coefficient of friction between the block and the surface is 0.3, find a) the initial velocity of the bullet, b) the velocity of the block right after being hit by the bullet, and c) the energy loss of the bullet during the collision
(b) Use the stopping distance of the block, X, to get its velocity after the bullet goes through, Vblock.
(1/2)M*Vblock^2 = M*Uk*g*X
(a) The momentum loss of the bullet equals the momentum gain of the block. Use that fact to calculate the initial bullet velocity, Vi.
(c) Bullet energy loss =
(1/2)Mbullet*(Vi^2 - 200^2)