SOMEONE PLEASE HELP IN PHYSICS!!?

Question

1) A 24g bullet strikes and becomes embedded in a 1.40kg block of wood placed on a horizontal surface just in front of the gun.

A) If the coefficient of kinetic friction between the block and the surface is 0.38, and the impact drives the block a distance of 9.5m before it comes to rest, what was the muzzle speed of the bullet? 

2) A 142g baseball moving 28.0m/s strikes a stationary 5.20kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.40m/s . 

A) What is the baseball's speed after the collision? 

B) Find the total kinetic energy before the collision. 

C) Find the total kinetic energy after the collision.

PLEASE EXPLAIN

Fireman · Accepted Answer

1) (A) Let the velocity of the bullet before the collision was v m/s and the velocity of the (bullet+block) is v m/s after the collision:-
=>By the work energy relation:-
=>W = ∆KE
=>Ff.s = 1/2mv^2
=>µk x N x s = 1/2(M+m)v^2
=>v^2 = [2µk x (M+m)g x s]/(M+m)
=>v = √[2µkgs]
=>v = √[2 x 0.38 x 9.8 x 9.5]
=>v = 8.41 m/s 
=>By the law of momentum conservation:-
=>m1u1+m2u2 = (m1+m2)v
=>24 x 10^-3 x u + 0 = (24 x 10^-3 + 1.40) x 8.41
=>u = 499.09 m/s
2) (A) By the law of momentum conservation:-
=>m1u1+m2u2 = m1v1+m2v2
=>142 x 10^-3 x 28 + 0 = 142 x 10^-3 x v1 + 5.20 x 1.40
=>v1 = -23.27 m/s [-ve just indicating that direction is opposite to initial velocity]
(B) KE(initial) = 1/2m1u1^2 = 1/2 x 142 x 10^-3 x (28)^2 = 55.66 J
(C) KE(final) = 1/2m1v1^2 + 1/2m2v2^2 = 1/2 x 142 x 10^-3 x (23.27)^2 + 1/2 x 5.20 x (1.4)^2 = 43.54 J

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SOMEONE PLEASE HELP IN PHYSICS!!?

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