a 1kg ball moving at 12m/s collides head on with a 2kg ball moving in the opposite direction with a speed of 24m/s.if the cofficient of resitution is 2/3, then find the energy lost in collision
Answers
m1 u1 + m2 u2 = m1 v1 + m2 v2 …............. [1]
where:
m1 = 1 kg
m2 = 2 kg
u1 = 12 m/s
u2 = -24 m/s
and the coefficient of restitution, e, is given by:
e = (v2 – v1)/(u1 – u2) …......................... [2]
a) e = 2/3
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 2/3 = (v2 – v1)/36
2/3 = (v2 – (-2 v2 – 36))/36
72/3 = 3 v2 + 36
► v2 = -12/3 = -4 m/s
► v1 = -2 v2 – 36 = -28 m/s
b)
[1] …............... m1 u1 + m2 u2 = (m1 + m2) v
1 * 12 + 2 * (-24) = (1 + 2) v
3 v = -36
v = -12 m/s
c) e = 1
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 1 = (v2 – v1)/36
1 = (v2 – (-2 v2 – 36))/36
36 = 3 v2 + 36
► v2 = 0
► v1 = -2 v2 – 36 = -36 m/s
Regards
where:
m1 = 1 kg
m2 = 2 kg
u1 = 12 m/s
u2 = -24 m/s
and the coefficient of restitution, e, is given by: e = (v2 – v1)/(u1 – u2) …......................... [2]
a) e = 2/3
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 2/3 = (v2 – v1)/36
2/3 = (v2 – (-2 v2 – 36))/36
72/3 = 3 v2 + 36
► v2 = -12/3 = -4 m/s
► v1 = -2 v2 – 36 = -28 m/
b)
[1] …............... m1 u1 + m2 u2 = (m1 + m2) v
1 * 12 + 2 * (-24) = (1 + 2) v
3 v = -36
v = -12 m/s
c) e = 1
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 1 = (v2 – v1)/36
1 = (v2 – (-2 v2 – 36))/36
36 = 3 v2 + 36
► v2 = 0
► v1 = -2 v2 – 36 = -36 m/s
Hence, the velocity of ball of mass 1 kg after collision is -28 m/s and ball with mass 2 kg, - 4m/s. This means, both the balls move in same direction after collision.