A 8.6 g bullet?

Question

is fired into a 122 g block of wood which is at rest on a horizontal surface and stays in the wood. After the impact, the block slides 14 m and then comes to rest. If the coefficient of friction between the surface and the block is 0.5, what is the speed of the bullet before impact. Acceleration of gravity is 9.8 m/s^2, answer in units of m/s.

Answer 1

It was a long time that I solved problems like these. I guess the following is correct:

1) Friction force in block:

f = mu*W = (0.5)(0.122 kg)(9.81 m/s²) = 0.598 N

the same value of friction factor we have for desacceleration of block:

a = 0.5 m/s²

2)We can use the value of displacement and desacceleration to know speed of sliding of wood block:

v² = 2ad = 2(0.5 m/s²)(14 m) = 14 m²/s²
v = sqrt (14) = 3.74 m/s
(I ommited negative sign for desacceleration because it just signals movement direction not absolute magnitude)

3)Assuming that collision was elastic:

m1v1 + m2v2 = m1' v1' + m2' v2' ....conservation of momentum

where: m1 =mass of bullet, m2 =mass of wood block
............v1, v2 = velocities before impact
............v1', v2' = velocities after impact

(8.6 g)(v1) + (122 g)(0 m/s) = (8.6 g)(0 m/s) + (122g)(3.74 m/s)
8.6 v1 +0 = 0 + 456.28 g m/s

We solve for v1:
v1 = 456.8 / 8.6 = 53.05 m/s

So, speed of bullet before impact is 53.05 m/s

Hope it helps!

Good luck!