Question

A 20 g bullet strikes a 1.5 kg wood block and becomes embedded in it (all horizontal motion). The coefficient of friction between the block and the surface is 0.3 and the impact makes the block move a distance of 7 m before it comes to rest. What was the muzzle speed (in m/s) of the bullet? Report your answer with two significant figures.

Answer 1

Step-by-step explanation:

Kinetic frictional force acting on the block = μ mg

where μ is coefficient of friction , m is mass of block.

.3 x 1.5 x 9.8 = 4.41 N .

Let v be the velocity of bullet + block after collision

kinetic energy of composite mass after the strike

= 1 /2 x 1.52 x v²

this will be equal to work done by friction .

.76 v² = 4.41 x 7

v² = 40.62

v = 6.37 m /s

Now we can obtain muzzle speed of bullet by applying conservation of momentum .

Let this speed be u

initial momentum of bullet

= .02 x u

final momentum of composite mass

= 1.52 x 6.37

.02 u = 1.52 x 6.37

u = 484.12 m /s .

= 480 m /s ( in two significant figures )