A 0.060kg tennis ball, moving with a speed of 5.40m/s , has a headon collision with a 0.090kg ball initially moving in the same direction at a speed of 3.38m/s . Assume that the collision is perfectly elastic.
Part A
Determine the speed of the 0.060kg ball after the collision.
Part B
Determine the direction of the velocity of the 0.060 ball after the collision.
in the direction of the initial velocity  
in the direction opposite to the initial velocity Part C Determine the speed of the 0.090kg ball after the collision. Part D Determine the direction of the velocity of the 0.090 ball after the collision.

according to conservation of linear momentum
m1*u1 + m2*u2 = m1*v1+m2*v2
m1*(u1v1) = m2*(v2u2)........(1)
according to conservation of energy
0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2
m1*(u1^2  v1^2) = m2*(v2^2u2^2).....(2)
from 1 &2
u1 + v1 = u2+v2
u1  u2 = v2  v1
v2 = u1  u2 + v1........(3)
3 in 1
m1*(u1v1) = m2*(u1  u2 + v1  u2)
v1 = u1*(m1m2)/(m1+m2) + 2*m2*u2/(m1+m2)
v2 = u2*(m2m1)/(m1+m2) + 2*m1*u1/(m1+m2)
m1 = 0.06 kg...........m2 = 0.09 kg
u1 = +5.4...........u2 = +3.38
v1 = ?...........v2 = ?
A) v1 = (((0.060.09)*5.4)/(0.06+0.09)) +
((2*0.09*3.38)/(0.06+0.09)) = + 2.976 m/s
B) in the direction of the initial
velocity
C) v2 = (((0.090.06)*3.38)/(0.06+0.09)) +
((2*0.06*5.4)/(0.06+0.09)) = +4.996 m/s
D) in the direction of the initial velocity
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