A spherical metal ball of mass 0.5kg moving with a speed of 0.5ms^-1 on a smooth linear horizontal track collides head on with another ball B of same mass at rest. Assuming the collision to be perfectly elastic,what are the speeds of A and B after collision?
Answers
momentum before = .5 * .5 + .5 * 0 = .25
energy before = (1/2) (.5) .5^2 = (1/2) .5^3/2
momentum after = .5 * v1 + .5 V2 = .25 still
energy after = (1/2) .5 v1^2+(1/2)(.5) v2^2=(1/2)(.5)(v1^2+v2^2)
=(1/2)(.5)^3/2 still
so v1+v2 = .5 so v2=.5-v1
and
v1^2+ v2^2 = .5^1/2
v1^2 + .25 - v1 + v1^2 = .25
2 v1^2 - v1 = 0
v1 (2v1-1) =0
either v1 = 0 or v1 = 0.5 after crashing through ball 2 without losing any speed
v1 = 0 and v2 = .5 -0 = .5
the first one stops, the second one goes on at original speed.
Given: A spherical metal ball of mass 0.5kg moving with a speed of 0.5m/s on a smooth linear horizontal track collides head on with another ball B of same mass at rest. The collision is elastic.
To find: Speeds of A and B after collision
Explanation: Initial velocity of ball A(u1)= 0.5 m/s
Mass of ball A(m1)= 0.5 kg
Final velocity of ball B(u2) = 0
Mass of ball B(m2) = 0.5 kg
Momentum before collision
= m1* u1 + m2* u2
=0.5* 0.5 + 0*0.5
= 0.25
Let final velocity of ball A be v1 and ball B be v2.
Momentum after collision= m1*v1 + m2* v2
= 0.5(v1 + v2)
Since it is an elastic collision, momentum remains constant.
0.5(v1+v2) = 0.25
=> v1 + v2 = 0.5
The coefficient of restitution of elastic collision is 1. Therefore,
Velocity of approach= Velocity of separation
=>u1 = v2-v1
=> v2 - v1 = 0.5
Adding v1+v2= 0.5 and v2-v1 = 0.5 we get:
2 v2 = 1
=> v2 = 0.5
=> v1 = 0
Therefore, the speed of ball A after collision is 0 and velocity of ball B after collision is 0.5 m/s.