Momentum and Impulse with friction?

Question

A 6g bullet is fired horizontally into a 2.8 kg block resting on a horizontal surface with a coefficient of friction of .3. When the bullet comes to rest in the block, the block has slid .65m. What is the velocity of the bullet?

the answer is 915 m/s but i don't know how to solve the problem.

Answer 1

The momentum of the bullet before striking the block is p = mV; where m = .006 kg and V is the velocity you are looking for.

Thus, mV = (m + M)v; where M is the mass of the block = 2.8 kg and v is the after impact velocity of the block with the embedded bullet. v is the initial velocity of the combined masses; so the u^2 = v^2 + 2ad; where u = 0, the end velocity after the masses slide d = .65 meters.

Thus, the acceleration (deceleration) of the combined masses is a = v^2/2d and the force on the combined masses is F = (m + M)v^2/2d = k(m + M)g; where k = .3. Therefore, we have v^2/2d = kg and v = sqrt(2kgd); where k = .3, g = 9.81 m/sec^2, and d = .65 meters.

Then from the conservation of momentum, we have mV = (m + M)v; so that V = [(m + M)/m] sqrt(2kgd), where m = .006 kg, M = 2.8 kg, k = .3, g = 9.81, and d = .65 so you have everything to work the math. [NB: I worked it and got 914.75 mps.]

Answer 2

least confusing physique of strategies could be in accordance with acceleration. because kinetic coefficient of friction is .25 the sled would be decelerating at .25g = 0.25*9.eighty one = 2.40 5 m/s^2. The time required for the sled to decelerate by ability of 3m/s is then: t = (substitute in v)/(acceleration) = 3/2.40 5 = a million.22 sec.