A bomb of mass 9kg initially at rest explodes into two pieces of masses 3kg and 6kg. If the kinetic energyof 3kg mass is 216J , then the velocity of 6kg mass will be
a) 4m/s
b) 3ms
c) 2m/s
d) 6m/s
Answers
Answered by
21
By law of conservation of momentum,
Total initial momentum of system = Total final momentum of system.
0 = m1v1 + m2v2
0 = 3*v1 + 6*v2
Since KE of 3kg = 216
=) 1/2 * 3*v1^2 = 216
=) v1^2 = 144
=) v1= 12ms-1.
Hence 0 = 3*12 + 6*v2
-36 = 6v2
=) - 6m/s = v2.
Velocity of 6 kg is in opposite direction of 3kg.
Hence velocity of 6kg mass will be 6m/s.
Hope it's helpful to u.
Total initial momentum of system = Total final momentum of system.
0 = m1v1 + m2v2
0 = 3*v1 + 6*v2
Since KE of 3kg = 216
=) 1/2 * 3*v1^2 = 216
=) v1^2 = 144
=) v1= 12ms-1.
Hence 0 = 3*12 + 6*v2
-36 = 6v2
=) - 6m/s = v2.
Velocity of 6 kg is in opposite direction of 3kg.
Hence velocity of 6kg mass will be 6m/s.
Hope it's helpful to u.
Answered by
16
From law of conservation of momentum,
> m1v1 = m2v2
> 3×v1 = 6×v2
> v1 = 2v2 .....(1)
ATQ , 1/2×3×v1² = 216
v1² = 216×2/3=144
Therefore, v1 = 12
From Eq. (1),
v2= v1/2 = 6
Answer is 6m/s
> m1v1 = m2v2
> 3×v1 = 6×v2
> v1 = 2v2 .....(1)
ATQ , 1/2×3×v1² = 216
v1² = 216×2/3=144
Therefore, v1 = 12
From Eq. (1),
v2= v1/2 = 6
Answer is 6m/s
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