Question

A 100 g bullet traveling in the x-direction at 100 m/s strikes a 1 kg wooden block at
rest. After the collision, wooden block splits into two parts [.2 kg and .8 kg] and the
bullet is observed traveling at a speed of 50 m/s in the x-direction. Assume that the
wooden pieces are traveling in the x-y plane and the .8 kg piece is traveling 30 degrees to the
right of the x-direction. If the kinetic energies of the wooden pieces are equal after the
collision,
i) In which direction is the .2 kg piece moving?
ii) What are the velocities of the two wooden pieces?
iii) How much kinetic energy is lost during the collision?

Answer 1

m1 = mass of first piece = 0.2 kg

m2 = mass of second piece = 0.8 kg

m3 = mass of bullet = 0.1 kg

before collision

v1i = v2i = 0           v3i = 100 i m/s

initial momentum Pbefore = m1*v1i + m2*v2i + m3*v3i = 0.1*100 = 10 i

after collision

K1 = K2

(1/2)*m1*v1^2 = (1/2)*m2*v2^2

0.2*v1^2 = 0.8*v2^2

v1 = 2*v2

v1f = v1x i + v1y j = v1*costheta i + v1*sinthetaj

v1f = 2*v2*costheta i + 2*v2*sintheta j

v2f = v2*cos30 i - v2*sin30 j  

v3f = 50 i

final momentum

Pafter = m1*v1f + m2*v2f + m3*v3f

Pafter = 0.2*2*v2*costheta i + 0.2*2*v2*sintheta j + 0.8*v2*cos30i - 0.8*v2*sin30 j + 0.1*50 i

Pafter = (0.4*costheta*v2 + 0.693*v2 + 5 ) i + (0.4*sintheta - 0.4)*v2 j

Pafter = Pbefore

10 i = (0.4*costheta*v2 + 0.693*v2 + 5 ) i + (0.4*sintheta - 0.4)*v2 j

(0.4*sintheta - 0.4)*v2 = 0

0.4*sintheta = 0.4

sintheta = 1

theta = 90

(a)

the 0.2 kg mass moves along +y direction

===================

(b)

10 = (0.4*costheta*v2 + 0.693*v2 + 5 )

10 = (0.4*cos90*v2 + 0.693*v2 + 5 )

v2 = 7.21 m/s

v1 = 2*v2 = 14.42 m/s

==============================

(c)

before collsion

Ki = (1/2)*m3*v3i^2 = (1/2)*0.1*100^2 = 500 J

after collision

K1 = (1/2)*m1*v1^2 = (1/2)*0.2*14.42^2 = 20.8 J

K2 = (1/2)*m2*v2^2 = (1/2)*0.8*7.21^2 = 20.8 J

K3 = (1/2)*m3*v3^2 = (1/2)*0.1*50^2 = 125 J

Kf = K1 + K2 + k3 = 166.6 J

Kf < ki

energy loss = Ki - Kf = 333.4 J