Question

In a ballistics test, a 20.0 gg bullet traveling horizontally at 1100 m/sm/s goes through a 30.0 cmcm -thick 450 kgkg stationary target and emerges with a speed of 900 m/sm/s . The target is free to slide on a smooth horizontal surface.

Part A

How long is the bullet in the target?

Part B

What average force does the bullet exert on the target?

Part C

What is the target's speed just after the bullet emerges?

Answer 1

Answer:

Given, mass of the bullet m = 20 g = 0.020 kg, initial speed of the bullet vi = 1100 m/s and final speed of the bullet vf = 900 m/s, distance travelled by the bullet inside the target is d = 30 cm = 0.30 m and mass of the target M = 450 kg.

(A) Average speed of the bullet is vavg = (vi + vf) /2 = (1100 m/s+900m/s)/2 = 1000 m/s

Time travelled by the bullet inside the target is t = d/vavg = 0.30 m/(1000 m/s) = 3.0 x 10-4 s.

(B) Change in momemtum of the target is pt and change in momentum of the bullet pb

According to conservation of momentum, pt = -pb = -m vb, where vb is the change in speed of the bullet and its value is vb = vf - vi = 900 m/s - 1100 m/s = -200 m/s, here '-' sign indicated the loss in the speed Therefore, pb = m vb = (0.020 kg) (200 m/s) = 4.0 kg m/s

The average force Favg that exeted by the bullet on the target is

Favg = pb/t = (4.0)/ (3.0x10-4) N = 1.3 x 104 N.

(C) Momentum of the target is pt = Mvt, where vt is the target's speed just after the buller emerges.

Therefore, vt = pt / M = pb / M = (4.0 kg.m/s)/450 kg = 8.89 x 10-3 m/s.