Asked by Cathy bens🐈
1.A 4kg ball moving at 8ms-1 collides with a stationary ball of mass 12kg,and they stick together.calculate the final velocity and kinetic energy lost in the impact.
2.when a person fires a rifle it is advisable to hold the butt firmly against the shoulder rather than a little way from it to minimise the impact on the shoulder.explain why?
2.when a person fires a rifle it is advisable to hold the butt firmly against the shoulder rather than a little way from it to minimise the impact on the shoulder.explain why?
Answers
Answered by
henry2,
Given:
M1 = 4 kg, V1 = 8 m/s.
M2 = 12 kg, V2 = 0.
V3 = velocity of M1 and M2 after the collision.
1. Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V3.
4*8 + 12*0 = 4*V3 + 12*V3,
V3 = 2 m/s.
KE before the collision:
0.5M1*V1^2 + 0.5M2*V2^2 = 2*8^2 + 6*0^2 = 128 + 0 = 128 Joules.
KE after the collision:
0.5M1*V3^2 + 0.5M2*V3^2 = 2*2^2 + 6*2^2 = 8 + 24 = 32 Joules.
KE Lost = 128-32 =
M1 = 4 kg, V1 = 8 m/s.
M2 = 12 kg, V2 = 0.
V3 = velocity of M1 and M2 after the collision.
1. Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V3.
4*8 + 12*0 = 4*V3 + 12*V3,
V3 = 2 m/s.
KE before the collision:
0.5M1*V1^2 + 0.5M2*V2^2 = 2*8^2 + 6*0^2 = 128 + 0 = 128 Joules.
KE after the collision:
0.5M1*V3^2 + 0.5M2*V3^2 = 2*2^2 + 6*2^2 = 8 + 24 = 32 Joules.
KE Lost = 128-32 =
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