Question

A 12 kg red ball travels with a velocity of 7 m/s towards a 8 kg blue ball that is traveling a a speed of 10 m/s in the opposite direction.What is the velocity of the blue ball after collision if the velocity of the red ball after collision is 5 m/s?(The balls have changed direction.)

Answer 1

A $12\mathrm{kg}$ red ball travels with a velocity of $7\mathrm{m / s}$ towards a $8\mathrm{kg}$ blue ball that is traveling a speed of $10\mathrm{m / s}$ in the opposite direction. What is the velocity of the blue ball after collision if the velocity of the red ball after collision is $5\mathrm{m / s}$ ?(The balls have changed direction.)

Solution:

Set OX as positive direction.

We are given:

$$m _ {r e d} = 1 2 k g$$ $$m _ {b l u e} = 8 k g$$ $$v _ {r e d 1} = 7 \frac {m}{s}$$ $$v _ {b l u e 1} = - 1 0 \frac {m}{s}$$ $$v _ {r e d 2} = - 5 \frac {m}{s}$$

According to the linear momentum conservation principle:

$$m _ {r e d} * v _ {r e d 1} + m _ {b l u e} * v _ {b l u e 1} = m _ {r e d} * v _ {r e d 2} + m _ {b l u e} * v _ {b l u e 2}$$

Thus:

$$v _ {b l u e 2} = \frac {m _ {r e d} * v _ {r e d 1} + m _ {b l u e} * v _ {b l u e 1} - m _ {r e d} * v _ {r e d 2}}{m _ {b l u e}}$$

Calculating:

$$v _ {b l u e 2} = \frac {1 2 * 7 + 8 * (- 1 0) - 1 2 * (- 5)}{8} = \mathbf {8} \frac {\mathbf {m}}{\mathbf {s}}$$

Answer: 8 m/s