Question

A 0.6-kg brick is thrown into a 25-kg wagon which is initially at rest. If, upon entering, the brick has a velocity of 10 m/s as shown, determine the final velocity of the wagon.

 

Answer 1

Concepts and reason

This question is based on the concept of momentum conservation. The law of conservation of momentum states that if no external force is acting on the system the momentum of the system remains conserved. Therefore, the initial momentum becomes equal to the final momentum. First, determine the velocity of the brick in the x-direction and then apply the law of conservation of momentum in the same direction to determine the velocity of the wagon.

Fundamentals

The expression of the momentum is as follows, \(P=m v\)

Here, \(P\) is the momentum particle, \(m\) is the mass particle and \(v\) is the velocity of the particle. The expression of the conservation of momentum is as follows,

\(\sum P_{\text {final }}=\sum P_{\text {initial }}\) Here, \(P_{\text {initial }}\) is the initial momentum, \(P_{\text {final }}\) is the final momentum.

The velocity of the wagon is determined as follows:

The initial velocity of brick makes an angle of \(30^{\circ}\) with horizontal therefore, the initial velocity of the brick in the \(\mathrm{x}\) -direction is:

\(V_{1 B, x}=V_{1 B} \cos \left(30^{\circ}\right) \ldots \ldots\) (1)

Here, \(V_{1 B, x}\) is the initial velocity of the brick in the \(\mathrm{x}\) -direction and \(V_{1 B}\) is the initial velocity of the brick. Substitute, \(10 \mathrm{~m} / \mathrm{s}\) for \(V_{1 B}\) in equation (1),

\(V_{1 B, x}=(10 \mathrm{~m} / \mathrm{s}) \cos \left(30^{\circ}\right) \mathrm{m} / \mathrm{s}\)

\(=8.66 \mathrm{~m} / \mathrm{s}\)

Initial velocity of the brick is \(30^{\circ}\) from the \(x\) -axis. So, take the cosine component of the velocity to get the velocity along \(x\) axis. The velocity is known so, the velocity of the Brick along \(x\) -axis can be calculated.

Apply the law of conservation of momentum in the \(\mathrm{x}\) -direction. \(m_{B} V_{1 B, x}+m_{W} V_{1 W, x}=m_{B} V_{2 B, x}+m_{W} V_{2 W, x} \ldots \ldots\) (2)

Here, \(m_{B}\) is the mass of the brick, \(V_{1 W, x}\) is the initial velocity of the wagon in the \(\mathrm{x}\) -direction, \(m_{W}\) is the mass of the wagon, \(V_{2 B, x}\) is the final velocity of the brick in the \(x\) -direction and \(V_{2 W, x}\) is the final velocity of the wagon in the \(x\) direction. After the brick has entered the wagon they travel with the same velocity in the x-direction, therefore, the final velocity of the brick becomes equal to the final velocity of the wagon. Therefore, \(V_{2 B, x}=V_{2 W, x}\). Initially, the wagon is stationary therefore its velocity is zero. Thus, \(V_{1 W, x}=0 \mathrm{~m} / \mathrm{s}\). Substitute, \(0.6 \mathrm{~kg}\) for \(m_{B}, 25 \mathrm{~kg}\) for \(m_{W}, 0 \mathrm{~m} / \mathrm{s}\) for \(V_{1 W, x}, 8.66 \mathrm{~m} / \mathrm{s}\) for \(V_{1 B, x}\) and \(V_{2 W, x}\) for \(V_{2 B, x}\) in equation (2),

\(0.6(8.66)+25(0)=0.6 V_{2 W, x}+25 V_{2 W, x}\)

\(5.196=25.6 \mathrm{~V}_{2 W, x}\)

\(V_{2 W, x}=\frac{5.196}{25.6}\)

\(=0.202 \mathrm{~m} / \mathrm{s}\)

The momentum of any system is conserved only when there is no external force acting on it, therefore, the initial momentum of the system is equated to the final momentum. The velocity of the brick becomes equal to the velocity of the wagon upon entering.

The final velocity of the wagon is \(0.202 \mathrm{~m} / \mathrm{s}\).