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Physics Question 10 – NEET-UG 2025

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A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g=9.8 m/s2)

Impulse is defined as the change in momentum of an object.

Video Walkthrough
Step 1: Calculate velocities before and after collision✦ Active

First, calculate the velocity of the ball just before it hits the ground (v1) using the initial height h1=40 m. Since it's dropped, the initial velocity is 0. Using the kinematic equation v2=u2+2gh, we get v1=2gh1.

v1=2×9.8 m/s2×40 m=784 m/s=28 m/s (downwards)

Next, calculate the velocity of the ball just after it rebounds from the ground (v2) using the rebound height h2=10 m. At the peak of its rebound, its final velocity is 0. Using v2=u2+2gh, we get v2=2gh2.

v2=2×9.8 m/s2×10 m=196 m/s=14 m/s (upwards)
💡 Teacher's Secret Hint

Ensure you use the correct height for each velocity calculation. g is positive when calculating speed due to fall or speed needed to reach a height.

Step 2: Calculate the change in momentum (Impulse)○ Expand

Impulse (J) is the change in momentum, J=Δp=m(vfvi). It's crucial to consider the direction of velocities. Let's take the upward direction as positive and the downward direction as negative.

Initial velocity (before collision) vi=28 m/s (downwards).

Final velocity (after collision) vf=+14 m/s (upwards).

J=m(vfvi)=0.5 kg×(14 m/s(28 m/s))
J=0.5 kg×(14+28) m/s=0.5 kg×42 m/s
J=21 NS
💡 Teacher's Secret Hint

The most common mistake is forgetting the vector nature of velocity and momentum. Always assign a consistent sign convention for directions.

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