Was Momentum Conservation Used Correctly in Calculating Bullet Velocity?

[Chocolaty]

At one time, the velocity of a rifle bullet was measured using a ballistic pendulum made up of a wooden block suspended from a string.

The block has a mass of 20kg. A 50g rifle bullet is fired into the block, penetrating it by 10cm and causing it to swing upwards, increasing the height by 20cm.

a) Calculate the velocity of the bullet as it enters the wooden block. Express your answer in Km/h

Ep = 20*9.8*0.2 = 39.2 J
Ex = 0.5mv^2
39.2 = 0.5*20*v^2
v = 2m/s

m1v1 + m2v2 = (m1 + m2)*v'
0.05*v1 = (20.05)*2
v1 = 802m/s = 2887.2 km/h

b) Calculate the frictional force between the bullet and the wooden block

Wf = Ff * d
39.2 J = Ff * 0.1
Ff = 392 N

I'm not sure i was allowed to use the m1v1 + m2v2 = (m1 + m2)*v' in a).
Can anyone confirm this?

 

[Chocolaty]

Anyone?
_________

 

[Chi Meson]

That is exactly what you do in part a. It is a collision; conservation of momentum governs the velocity outcomes of collisions. Work-energy theorem provides answers to the "force-distance" problems, and impulse-momentum theorem works for the "force-time" problems.