A75 kgman rides on a39 kgcart moving at a velocity of 2.3 m/s. He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart’s velocity, including sign?

1. Mass of the man,$m_1=75\mathrm{kg}$.
2. Mass of the cart,$m_2=39\mathrm{kg}$.
3. The velocity of the cart with man,$v_{1i}=2.3\mathrm{m}/\mathrm{s}$.
4. The velocity of the man after the jump, role="math" localid="1661252311079" $v_{1f}=0\mathrm{m}/\mathrm{s}$.

Using the formula of conservation of momentum, we can find the final speed of the cart. Then using this speed, we can find the resulting change in the cart’s velocity, including the sign.

$$\begin{gathered} \left({\mathrm{m}}_1+{\mathrm{m}}_2\right)v_{1i}=m_1{v}_{1f}+m_2{v}_{2f} \\ ∆v=v_{2f}-v_{1i} \end{gathered}$$

Here $v_{2f}$ is the velocity of the cart after the man jumps off.

For conservation of momentum, we can write the momentum equation as,

$\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)v_{1i}=m_1{v}_{1f}+m_2{v}_{2f}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \left(75\mathrm{kg}+39\mathrm{kg}\right)\left(2.3\frac{\mathrm{m}}{\mathrm{s}}\right)=\left(75\mathrm{kg}\right)\left(0\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(39\mathrm{kg}\right)v_{2f} \\ \left(262.2\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)=\left(39\mathrm{kg}\right)v_{2f} \\ {v}_{2f}=\frac{\left(262.2\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)}{\left(39\mathrm{kg}\right)} \\ =6.7\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

So, change in the velocity can be calculated as,

$∆v=v_{2f}-v_{1i}$

Substitute the values in the above expression, and we get,

$∆v=6.7\frac{\mathrm{m}}{\mathrm{s}}-2.3\frac{\mathrm{m}}{\mathrm{s}}$

$∆v=+4.4\frac{\mathrm{m}}{\mathrm{s}}$

Therefore, the resulting change in the cart’s velocity, including the sign, is +4.4 m/s .