Question

A heavy bullet of mass m = 0.1200 kg is fired with an initial velocity v = 400. m/s into a large block of lightweight wood of mass M = 0.500 kg, which is initially at rest on a frictionless surface. This initial situation is depicted on the left side of the picture below. The bullet gets stuck inside of the wood block and both move together thereafter as shown in the picture.
a) What is the velocity V of the wood block and bullet system right after the perforation?
b) A small time after the perforation is complete the block (w/bullet inside) slides onto a rough surface with a coefficient of kinetic friction μk = 0.250. It comes to a stop after sliding a distance ∆x on the rough surface as shown on the right side of the picture above. What is the value of ∆x?

Answer 1

(a)

This is an inelastic collision

let bullet be denoted as 1

let block be denoted as 2

we have

m1v1i + m2v2i = ( m1 + m2)v

0.12 * 400 + 0 = ( 0.12 + 0.5) v

solve for v, we get

v = 77.42 m/s

__________________________

(b)

Now, we can use kinematics

first, we find the deceleration

Fnet = ma

umg = ma

a = - ug

a = - 0.25 * 9.8

a = -2.45 m/s²

so,

v² - u² = 2a$\Delta$x

where v = 0 = final velocity

u = 77.42 m/s

so,

0² - 77.42² = 2 * - 2.45 * $\Delta$ x

so,

$\Delta$x = 1223.2 m