Physics collisions?

(a) conserve momentum

0.0210kg * 450m/s + 0 = 0.5210kg * V

solves to

V = 18.1 m/s

but you apparently knew that.

(b) initial KE = ½ * 0.521kg * (18.1m/s)² = 85.7 J

friction work W = µmgd = 0.3 * 0.521kg * 9.8m/s² * 8.0m = 12.3 J

and so the remaining KE is

KE' = (85.7 - 12.3) J = 73.4 J = ½ * 0.521kg * v²

and so the new velocity is

v = 16.8 m/s

which you apparently also knew.

(c) conserve momentum again:

0.521kg * 16.8m/s + 0 = 2.521kg * U

U = 3.47 m/s

so new KE is

KE" = ½ * 2.521kg * (3.47m/s)² = 15.2 J

What distance is required to do this much friction work?

15.2 J = 0.3 * 2.521kg * 9.8m/s² * d

solves to

d = 2.05 m

Hope this helps!

Momentum is conserved in a collision.

Initial momentum = 0.0210 * 450 = 9.45 kg * m/s

Total mass = 0.0210 + 0.5 = 0.521 kg

Final momentum = 0.521 * v

0.521 * v = 9.45

v = 9.45 ÷ 0.521

Their velocity is approximately 18.1 m/s

\(b)

The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity (in m/s)?

The friction force causes the bullet-embedded block to decelerate.

Ff = -0.3 * 0.521 * 9.8 = -1.53174 N

a = -1.53174 ÷ 0.521 = -2.94 m/s^2

Let’s use the following equation to determine its velocity.

vf^2 = vi^2 + 2 * a * d

vf^2 = (9.45 ÷ 0.521)^2 + 2 * -2.94 * 8

vf = √[((9.45 ÷ 0.521)^2 – 47.04]

Their velocity is approximately 16.8 m/s.

The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far (in m) does this combination travel before stopping?

Initial momentum = 9.45 kg * m/s

Total mass = 0.521 + 2 = 2.521 kg

Final momentum = 2.521 * v

2.521 * v = 9.45

v = 9.45 ÷ 2.521

Their velocity is approximately 3.75 m/s. The acceleration is still -2.94 m/s^2. Let’s use the following equation to determine the distance they move.

vf^2 = vi^2 + 2 * a * d

0 = (9.45 ÷ 2.521)^2 + 2 * -2.94 * d

5.88 * d = (9.45 ÷ 2.521)^2

d = (9.45 ÷ 2.521)^2 ÷ 5.88

The distance is approximately 2.39 meters. I hope this is helpful for you.