Question

Water leaves a hose at a rate of 3.5-kg per 2.9-s with a speed of 15.6-m/s and is aimed at a toy car at the exact moment the car is hit the water slows down to 4.7-m/s (That is, we ignore any splashing back.)

What is the force exerted by the water on the toy car?

Answer 1

Answer:

Assume that by "speed" the problem refers to the horizontal speed of the water jet. The water jet would exert approximately $\rm 13\; N$ of force on the toy car.

Explanation:

By Newton's Second Law, the net force on an object is proportional to the rate of change in its momentum. The momentum of an object is equal to its mass $m$ times its velocity $v$. $p = m \cdot v$.

The initial velocity of that $\rm 3.5\; kg$ of water is $\rm 15.6\; m \cdot s^{-1}$. It initial momentum would be:

$\begin{aligned}& p(\text{initial}) \\ &= m \cdot v(\text{initial}) \\ &= 3.5\; \rm kg \times 15.6 \; m \cdot s^{-1} = 54.60\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$.

The final velocity of that $\rm 3.5\; kg$ of water is $\rm 4.7\; m \cdot s^{-1}$. It final momentum would be:

$\begin{aligned}& p(\text{final}) \\ &= m \cdot v(\text{final}) \\ &= 3.5\; \rm kg \times 4.7 \; m \cdot s^{-1} = 16.45 \; \rm kg \cdot m \cdot s^{-1}\end{aligned}$.

The change in momentum (in the horizontal direction) is:

$\begin{aligned}& \Delta p \\ &= \rm 54.60\; kg \cdot m \cdot s^{-1} - 16.45\; \rm kg \cdot m \cdot s^{-1} \\ &= 38.15\; \rm kg \cdot m \cdot s^{-1} \end{aligned}$.

That took place in $2.9\; \rm s$. Assume that water leaves the hose at a constant rate. The (average) rate of change in the momentum of that water jet (in the horizontal direction) would be:

$\begin{aligned}& \frac{\Delta p}{\Delta t} \\ &= \frac{\rm 38.15 \; kg \cdot m \cdot s^{-1}}{2.9\; \rm s} \\ &\approx 13\; \rm kg \cdot m \cdot s^{-2} = 13\; \rm N\end{aligned}$.

Note, that one Newton is the same as $\rm 1\; kg \cdot m \cdot s^{-2}$.

By Newton's Second Law, the net force on the water in the horizontal direction would also be $\rm 13\; N$. Gravity acts in the vertical direction, so that would not be part of that force. The only possible source of that force would be the reaction force from the toy car.

By Newton's Third Law, the size of an action force and its reaction force should be the same. As a result, the action force of the water jet on the car should also be approximately $\rm 13\; N$.