Question

A bullet of mass 24grams traveling in a horizontal path with a velocity of 450 m/s strikes a wooden block of mass 976 grams resting on a rough surface. After impact, the bullet and the block move together for a distance of 7.5m before coming to rest.

a)calculate the force which brings the two bodies to rest
b)determine the coefficient of friction between the block and the surface during motion

Answer 1

a) F= 324N

b) u = 518.4

Step-by-step explanation:

a/ 2 soft colliders: mv1^2 + mv2^2 = MV^2

<=> 0.024*450^2 +0.967*0^2=(0.024+0.976)V^2

<=>V= 69.71 m/s

v^2-v0^2= 2as

0^2 - (69.71)^2 = 2a*7.5

<=> a= -324m/s^2

We have: -F = ma (because F is resistance, it is negative)

-F= 1*-324

F= 324N

Therefore, the force that makes the two bodies at rest is 324N.

b) We have the formula: F pulling force - F friction = ma

<=> MV^2 - umg = ma

<=> 1*(69.71)^2 - u*10 = -324

<=> u = 518.4

So the coefficient of friction is 518.4