Question:
The ball initially has a velocity of 5 m/s in the positive x-direction and some time later has a velocity of 7 m/s in the positive y-direction. If the work done by the ball is 48 joule, then the mass of the ball in kg is:
The ball initially has a velocity of 5 m/s in the positive x-direction and some time later has a velocity of 7 m/s in the positive y-direction. If the work done by the ball is 48 joule, then the mass of the ball in kg is:
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Use the work-energy theorem: Work done = change in kinetic energy, and remember that velocities in perpendicular directions can be treated using their magnitudes when calculating energy change.
Updated On: Jun 19, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the problem.
We are given initial and final velocities of a ball in perpendicular directions, and the work done by the force is given. We are asked to find the mass of the ball. The work-energy theorem applies: \[ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \] where \(v_i\) and \(v_f\) are magnitudes of initial and final velocities.
Step 2: Calculating initial and final speeds.
- Initial velocity magnitude along x: \(v_i = 5~\text{m/s}\)
- Final velocity magnitude along y: \(v_f = 7~\text{m/s}\)
Step 3: Applying work-energy theorem.
\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \] Substitute \(W = 48~\text{J}, v_i = 5, v_f = 7\): \[ 48 = \frac{1}{2} m (7^2 - 5^2) \] \[ 48 = \frac{1}{2} m (49 - 25) \] \[ 48 = \frac{1}{2} m \cdot 24 \] \[ 48 = 12 m \]
Step 4: Solve for mass.
\[ m = \frac{48}{12} = 4~\text{kg} \]
Step 5: Conclusion.
The mass of the ball is \(4~\text{kg}\).
We are given initial and final velocities of a ball in perpendicular directions, and the work done by the force is given. We are asked to find the mass of the ball. The work-energy theorem applies: \[ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \] where \(v_i\) and \(v_f\) are magnitudes of initial and final velocities.
Step 2: Calculating initial and final speeds.
- Initial velocity magnitude along x: \(v_i = 5~\text{m/s}\)
- Final velocity magnitude along y: \(v_f = 7~\text{m/s}\)
Step 3: Applying work-energy theorem.
\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \] Substitute \(W = 48~\text{J}, v_i = 5, v_f = 7\): \[ 48 = \frac{1}{2} m (7^2 - 5^2) \] \[ 48 = \frac{1}{2} m (49 - 25) \] \[ 48 = \frac{1}{2} m \cdot 24 \] \[ 48 = 12 m \]
Step 4: Solve for mass.
\[ m = \frac{48}{12} = 4~\text{kg} \]
Step 5: Conclusion.
The mass of the ball is \(4~\text{kg}\).
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