Question

Bob, who has a mass of, can throw a rock with a speed of $30m/s$. The distance through which his hand moves as he accelerates the rock from rest until he releases it is $1.0m$.

a. What constant force must Bob exert on the rock to throw it with this speed?

b. If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

Short answer

(a) The force Bob must exert on the rock to throw is $225N$

(B) The recoil speed after releasing the rock is$0.2m/s$

Step-by-step solution

Given information (part a)

Mass of Bob,$m=75kg$

Weight of rock weight, $m=0.5kg$

Speed of the rock,$30m/s$

Explanation (part a)

Acceleration :

$$\begin{gathered} v^2=u^2+2as \\ \text{Initialvelocity}u=0 \\ a=\frac{v^2-u^2}{2s} \\ a=\frac{{\left(30m/s\right)}^2-0}{2\left(1m\right)} \\ a=450m/s^2 \end{gathered}$$

Force exerted on the rock

$$\begin{gathered} F=ma \\ F=\left(0.5kg\right)\left(450m/s^2\right) \\ F=225N \end{gathered}$$

Given information (part b)

Mass of Bob, $m=75kg$

Weight of rock weight, localid="1649688983057" $m=0.5kg$

Speed of the rock,$v=30m/s$

Explanation (part b)

By conservation of momentum

$$\begin{gathered} m_1{v}_1+m_2{v}_2=0 \\ \text{where}{m}_{1}and{v}_1\text{arethemassandvelocityoftherock,} \\ {m}_{2}and{v}_2\text{arethemassandrecoilvelocityofbob.} \\ \text{Sincethespeedofrockisoppositetobob'srecoilspeed}{v}_1=-30m/s \\ \left(0.5kg\right)\left(-30m/s\right)+\left(75kg\right)v_2=0 \\ {v}_2=0.2m/s \end{gathered}$$