# Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of –0.1 m/s. What is the resulting velocity of marble B after the collision?

2
dari xinrteriahmuhd

## Jawapan

• meran
• Bercita-cita tinggi
2016-03-14T20:34:26+08:00
Fomula=m₁u₁+m₂u₂=m₁v₁+m₂v₂
=(0.08×0.5)+(0.05×0)=(0.08×-0.1)+(0.05×α)
=0.04+0=-0.008+0.05α
=0.04+0.008=0.05α
=0.048=0.05α
=0.048÷0.05=α
=0.96=α
α=velocity marble B after collision
2016-03-15T20:57:13+08:00
If i remember any mechanics correctly from last year, it's that the resultant mass x velocity should be equal to them before the collision. So, 0.08 x 0.5 = 0.04 and 0.05 x 0 = 0.
0.04 + 0 = 0.08 x 0.1 + 0.05 x W.( I believe you can ignore negatives here as it just dictates direction)
0.04 = 0.008 + 0.05W
0.032 = 0.05W
W = 0.64
So the resulting velocity is, hopefully, 0.64.
(Don't hurt me if this is wrong, I had the best intentions =] )
Good luck, this part of maths is brutal.