Question

A 5.00 gg bullet is fired horizontally into a 1.20 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 mm along the surface before stopping.

Answer 1

V= 295.2m/s

Step-by-step explanation:

Given that mass of bullet = 5.0g = 0.005kg

Mass of wooden block = 1.20kg

The coefficient of kinetic friction between block and surface = 0.20

The force of friction of the block

Fk = (μ) mg

Fk = 0.2×1.2×9.8m/s

F = 2.352N

= Force × distance

The work done = 2.352× 0.390

= 0.91728J

The initial velocity of the block can be calculated

1/2mv^2

0.5×1.2×v^2 = 0.917J

0.6V^2 = 0.917J

V^2 = 0.917J/0.6

V^2 = 1.52

V = √1.52

V = 1.23m/s

Now we shall use conservation of momentum to find the velocity of the bullet

0.005kg×v = 1.2kg × 1.23m/s

0.005v = 1.476

V= 1.476/0.005

V= 295.2m/s