Question

A rocket sled with a mass of 2900 kg moves at $\mathbf{250}\frac{\mathbf{m}}{\mathbf{s}}$on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after 920 kgof water has been scooped up. Ignore any retarding force on the scoop.

Short answer

Speed of sled after 920 kg water is scooped up is $190\frac{\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Step 1: Given

1. Rocket sled mass is,$m_1=2900\mathrm{kg}$.
2. Speed of sled is,$v_1=250\frac{\mathrm{m}}{\mathrm{s}}$.
3. Mass of water is, $m_2=920\mathrm{kg}$.

Determine the formulas and solve as:

Formula is as follow:

$m_1{v}_1=(m_1+m_2)v_2$

Here, m₁, m₂ are masses, v₁, v₂ are velocities.

Determining the speed of sled after 920 kg water is scooped up

According to the law of conservation of momentum,

$$\begin{gathered} m_1{v}_1=(m_1+m_2)v_2 \\ 2900\times 250=(2900\times 920)\times v_2 \\ {v}_2=189.79\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

In two significant figure:

$v_2=190\frac{\mathrm{m}}{\mathrm{s}}$

This is the speed of the water which is equal to the speed of the sled.

Therefore, speed of the sled after 920 kg water is scooped up is$190\frac{\mathrm{m}}{\mathrm{s}}$

Using the law of conservation of momentum, the speed of one of the objects undergoing elastic collision can be found if the speed of the system after collision is known.