Question

A blue marble of mass 0.350 kg traveling at 2.60 m/s to the left collides with a red marble of mass 0.350 kg that is initially at rest. After the collision, the blue marble is at rest. What is the new velocity of the red marble that is now moving in the same direction?

A.2.60 m/s to the left
B.0.350 m/s to the right
C.5.20 m/s to the left
D.0.91 m/s to the left​

Answer 1

`2.60\; {\rm m\cdot s^(-1)}` to the left.

Step-by-step explanation:

When an object of mass
`m` travels at a velocity of
`v`, the momentum of that object would be
`p = m\, v`. Note that since velocity is a vector quantity (has a direction) while mass is a scalar, the direction of momentum would be the same as that of velocity.

Before the collision:

• Momentum of the blue marble:
  `(0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)})` to the left.
• Momentum of the red marble would be
  `0\; {\rm kg\cdot m\cdot s^(-1)}` since velocity was
  `0\; {\rm m\cdot s^(-1)}`.

Immediately after the collision:

• Momentum of the blue marble would be
  `0\; {\rm kg\cdot m\cdot s^(-1)}`.
• The momentum of the red marble needs to be found.

Momentum is conserved immediately before and after the collision. In other words, the total momentum immediately after the collision would be the same as that immediately before the collision.

In this example, total momentum was
`(0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)})` to the left immediately before the collision. Hence, the total momentum immediately after the collision would also be
`(0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)})\!` to the left.

Subtract the momentum of the blue marble (
`0\; {\rm kg\cdot m\cdot s^(-1)}`) from the total momentum to find the momentum of the red marble:

`\begin{aligned}& (0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)}) - 0\; {\rm kg\cdot m\cdot s^(-1)} \\ =\; & (0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)}) && (\text{to the left}) \end{aligned}`.

Divide momentum by mass to find velocity:

`\begin{aligned} v &= (p)/(m) \\ &= \frac{(0.350\; {\rm kg})\, (2.60\; {\rm m\cdot s^(-1)}) }{0.350\; {\rm kg}}&& \genfrac{}{}{0em}{}{(\text{to the left})}{} \\ &= 2.60\; {\rm m\cdot s^(-1)} && (\text{to the left})\end{aligned}`.

Therefore, the velocity of the red marble would be
`2.60\; {\rm m\cdot s^(-1)}` to the left immediately after the collision.