two balls of mass 50 gram 100 gram moving along the same line and direction with velocities 3 metre per second and 1.5 metres per second respectively they collide and after collision with the second ball moves with a velocity 1.75 metres per second calculate the velocity of the first ball
Answers
Answer:
2.5 m/s
Explanation:
Let First Ball Be Called = A ,
and Second Be = B
Before Collision
Mass of Ball A (ma) = 50 g = 0.05 kg
Initial Velocity of Ball A (ua) = 3 m/s
Mass of Ball B (mb) = 100g = 0.1 kg
Initial Velocity of Ball B (ub) = 1.5 m/s
Momentum Formula = m × v
Momentum Formula = m × vwhere m is mass and v is velocity respectively.
Total Initial momentum (Ip) = ma(ua) + mb(ub)
Ip = 0.05 × 3 + 0.1 × 1.5
Ip = 0.30
After Collision (mass remains same)
Final Velocity of Ball A (va) = ?
Final Velocity of Ball B (vb) = 1.75 m/s
Total Final Momentum (Fp) = ma(va)+ mb(vb)
Fp = 0.05 × va + 0.1 × 1.75
Fp = 0.05 × va + 0.175
As per the law of conservation,
Total Initial Momentum = Total Final Momentum
Ip = Fp
0.30 = 0.05va + 0.175
0.125 = 0.05va
0.125/0.5 = va
2.5 = va
va = 2.5 m/s
Hence, Velocity Of First Ball after collision is 2.5 m/s.