Question

A truck of mass \( M \) and a car of mass \( \frac{M}{10} \) moving with the same momentum are brought to halt by the application of the same breaking force. The ratio of the distances travelled by the truck and car before they come to stop is:

• A. \( 1:10 \)
• B. \( 1:\sqrt{10} \)
• C. \( 100:1 \)
• D. \( 5:1 \)

Hint:

The work done by a constant force is \( W = \vec{F} \cdot \vec{d} \). When the force opposes the motion, \( W = -Fd \). The kinetic energy of an object is \( KE = \frac{1}{2} mv^2 \).

Answer 1

The Correct Option is A

Step 1: Establish the relationship between velocities using equal momentum.
The momentum \( p \) of an object is given by \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. Given that the truck and the car have the same momentum: \[ p_{truck} = p_{car} \] \[ m_{truck} v_{truck} = m_{car} v_{car} \] Substituting the given masses \( m_{truck} = M \) and \( m_{car} = \frac{M}{10} \): \[ M v_{truck} = \frac{M}{10} v_{car} \] Solving for the velocity of the car \( v_{car} \) in terms of the velocity of the truck \( v_{truck} \): \[ v_{car} = 10 v_{truck} \]
Step 2: Apply the work-energy theorem to determine the stopping distances.
The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy \( (\Delta KE = W_{net}) \). Here, the breaking force \( F \) does work to bring the vehicles to a stop. The initial kinetic energy is \( KE_i = \frac{1}{2} mv^2 \) and the final kinetic energy is \( KE_f = 0 \). The work done by the constant braking force \( F \) over a distance \( d \) is \( W = -Fd \) (negative because the force opposes the displacement). For the truck: \[ \Delta KE_{truck} = KE_{f,truck} - KE_{i,truck} = 0 - \frac{1}{2} M v_{truck}^2 = -\frac{1}{2} M v_{truck}^2 \] \[ W_{truck} = -F d_{truck} \] By the work-energy theorem: \[ -F d_{truck} = -\frac{1}{2} M v_{truck}^2 \implies d_{truck} = \frac{M v_{truck}^2}{2F} \] For the car: \[ \Delta KE_{car} = KE_{f,car} - KE_{i,car} = 0 - \frac{1}{2} \left(\frac{M}{10}\right) v_{car}^2 \] Substitute \( v_{car} = 10 v_{truck} \): \[ \Delta KE_{car} = -\frac{1}{20} M (10 v_{truck})^2 = -\frac{1}{20} M (100 v_{truck}^2) = -5 M v_{truck}^2 \] \[ W_{car} = -F d_{car} \] By the work-energy theorem: \[ -F d_{car} = -5 M v_{truck}^2 \implies d_{car} = \frac{5 M v_{truck}^2}{F} \]
Step 3: Calculate the ratio of the stopping distances. The ratio of the distances travelled by the truck and the car is: \[ \frac{d_{truck}}{d_{car}} = \frac{\frac{M v_{truck}^2}{2F}}{\frac{5 M v_{truck}^2}{F}} = \frac{M v_{truck}^2}{2F} \times \frac{F}{5 M v_{truck}^2} = \frac{1}{2 \times 5} = \frac{1}{10} \] Thus, the ratio of the distances travelled by the truck and the car before they come to a stop is \( 1:10 \).