Question

Fred (m=67 kg) foolishly attempts top jump from a canoe (45kg) to the dock. if he jumps forward with a speed of 3.3 m/s how fast will the canoe move backward

Answer 1

Approximately
`4.9\; {\rm m\cdot s^(-1)}` backward, assuming that the canoe was initially stationary.

Step-by-step explanation:

When an object of mass
`m` is moving at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

With a mass of
`m = 67\; {\rm kg}` and a velocity of
`v = 3.3\; {\rm m\cdot s^(-1)}` forward, the momentum of Fred would be
`p = m\, v = 67\; {\rm kg} * 3.3\; {\rm m\cdot s^(-1)} = 22.11\; {\rm kg \cdot m \cdot s^(-1)}` (forward.)

Before the jump, the velocity of the boat and Fred were both
`0\; {\rm m \cdot s^(-1)}`. Hence, the momentum of both Fred and the boat would initially be
`0\; {\rm kg \cdot m \cdot s^(-1)}`.

Momentum is preserved immediately after the jump. Thus, the momentum of Fred plus the momentum of the boat would still be
`0\; {\rm kg \cdot m \cdot s^(-1)}` after the jump.

Since the momentum of Fred was
`22.11\; {\rm kg \cdot m \cdot s^(-1)}`, the momentum of the boat would be
`(-22.11\; {\rm kg \cdot m \cdot s^(-1)})` (backward, as the negative sign indicates.)

Rearrange
`p = m\, v` to find velocity in terms of mass
`m` and momentum
`p`:

`\begin{aligned} v &= (p)/(m) \\ &= \frac{-22.11\; {\rm kg \cdot m \cdot s^(-1)}}{67\; {\rm kg}} \\ &\approx (-4.9)\; {\rm m \cdot s^(-1)}\end{aligned}`.

In other words, the velocity of the boat would be approximate
`4.9\; {\rm m\cdot s^(-1)}` backwards.