m1=10kg, m2=20kg, v1i=6m/s , v2i=-4m/s
a) By law of conservation of momentum,
m1v1i+m2v2i=m1v1f+m2v2f ------------(1)
since inelastic collision,
m1v1i+m2v2i=(m1+m2)vf
10*6+20*-4 = (10+20)vf     =>
vf = -0.67m/s
b) By eqn (1),
10*6+20*-4 =10*-7+20*v2f     Â
=> v2f = 2.5m/s ………………directed to right
c) For perfectly elastic collision KE must be conserved.
1/2m1v1i^2 +1/2m2v2i^2 =1/2m1v1f^2+1/2m2v2f^2
1/2*10*6^2+1/2*20*-4^2 =1/2*10*7^2+1/2*20*2.5^2
340 ¹ 307.5
Thus this is not perfectly elastic
collision
d) F1 = m1(v1f-v1i)/Δt = 10(-7-6)/0.03 = -4333.3 N
……………..directed to left
F2= m2(v2f-v2i)/Δt = 20(2.5-(-4))/0.03 = 4333.3 N
……………..directed to right
e) By law of conservation of momentum,
m1v1ix+m2v2ix=m1v1fx+m2v2fx
10*6+20*-4 =10*-7+20*v2fx     =>
v2fx = 2.5m/s ………………directed to right
f) By law of conservation of momentum,
m1v1iy+m2v2iy=m1v1fy+m2v2fy
10*0+20*0 =10*0+20*v2fx     =>
v2fx = 0m/s
