Question

A ball of mass 0.16kg is moving forwards at a speed of 0.50 m / s. A second ball of mass 0.10 kg is stationary. The first ball strikes the second ball. The second ball moves forwards at a speed of 0.50 m / s.

What is the speed of the first ball after the collision?
I have a physics exam coming up, can someone please help me???

Answer 1

Answer:

Simply apply conservation of momentum as the net external force on system is zero.

0.16×0.50 + 0.10×0=0.10×0.50 + 0.16×v

0.080=0.050+0.16v

0.16v= 0.030

v= 0.375 m/s

Hope it helps.

Answer 2

SoluTion :-

Momentum before collision = Momentum after collision

$\sf {m_{1}u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}}$

$\sf {(0.16 kg) (0.50 m/s) + (0.10 kg) (0 m/s) = (0.16 kg) v_{1} + (0.10 kg) (0.50 m/s)}$

$\sf {0.08\ kg\ m/s = (0.16 kg) v_{1} + 0.05\ kg\ m/s}$

$\sf {0.03\ kg\ m/s = (0.16\ kg) v_{1}}$

$\sf {v_{1} = 0.19 m/s}$