Question

2. A 10 kg ball moving to the right with a speed of 6.0 m/s collides with a 20 kg ball moving to the left at a speed of 4.0 m/s.

(a) If this is a head-on, perfectly inelastic collision, what is the velocity of the balls after the collision. Give magnitude and direction. (5 pts.)

(b) If this is a head-on collision and the the 10 kg ball moves to the left with a speed of 7.0 m/s after the
collision, what is the velocity of the 20 kg ball after the collision? Give magnitude and direction.

(c) Is the collision in part b perfectly elastic? (You must show a calculation here.) (5 pts.)

(d) If the balls are in contact for 0.03 seconds during the collision in part (b), what is the magnitude of
the average force that each ball exerts on the other? (5 pts.)

(e) Write an equation for conservation of momentum in the x direction and determine the x component of the
velocity of the 20 kg mass. (5 pts.)

(f) Write an equation for conservation of momentum in the y direction and determine the y component of the

velocity of the 20 kg mass. (5 pts.)

Answer 1

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       => 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