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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A 100 g100 \text{ g} iron ball having velocity 10 m/s10 \text{ m/s} collides with a wall at an angle 3030^\circ and rebounds with the same angle. If the period of contact between the ball and wall is 0.1 second0.1 \text{ second}, then the force experienced by the wall is:

A

10 N10 \text{ N}

B

100 N100 \text{ N}

C

1.0 N1.0 \text{ N}

D

0.1 N0.1 \text{ N}

Step-by-Step Solution

Let the mass of the iron ball be m=100 g=0.1 kgm = 100 \text{ g} = 0.1 \text{ kg} and its velocity be v=10 m/sv = 10 \text{ m/s}. The ball collides with the wall at an angle θ=30\theta = 30^\circ with the wall. The component of velocity perpendicular to the wall before the collision is vsin30v \sin 30^\circ. Since the ball rebounds with the same angle, the component of velocity perpendicular to the wall after the collision is vsin30-v \sin 30^\circ (assuming the collision is elastic). The magnitude of the change in momentum of the ball along the normal is: Δp=m(vsin30(vsin30))=2mvsin30\Delta p = m(v \sin 30^\circ - (-v \sin 30^\circ)) = 2mv \sin 30^\circ Substituting the given values: Δp=2×0.1×10×sin30=2×1×0.5=1 kg m/s\Delta p = 2 \times 0.1 \times 10 \times \sin 30^\circ = 2 \times 1 \times 0.5 = 1 \text{ kg m/s} The force experienced by the wall is equal to the rate of change of momentum (according to Newton's second law of motion) : F=ΔpΔtF = \frac{\Delta p}{\Delta t} Given the period of contact Δt=0.1 s\Delta t = 0.1 \text{ s}: F=10.1=10 NF = \frac{1}{0.1} = 10 \text{ N} Thus, the force experienced by the wall is 10 N10 \text{ N}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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