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Question #42412

Difficulty: 8/10 | 🟢 Free

The Question

A small ball is projected from the origin $(0,0)$ over level ground with an initial velocity $\vec{u} = 30\hat{i} + 40\hat{j} \text{ m/s}$ . At the highest point of its trajectory, it strikes the smooth inclined face of a massive wedge. The wedge is moving with a constant velocity $\vec{v}_w = 10\hat{j} \text{ m/s}$ . The outward normal to the wedge's inclined face is $\hat{n} = \frac{-\hat{i} + \hat{j}}{\sqrt{2}}$ . If the coefficient of restitution for the collision is $e = 0.5$ , what is the maximum height from the ground reached by the ball? (Take $g = 10 \text{ m/s}^2$ )

Options

A.
$125 \text{ m}$
B.
$100 \text{ m}$
C.
$160 \text{ m}$
D.
$80 \text{ m}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Maximum Height Trap|Coefficient of Restitution|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle impulsive non-conservative forces during collisions in moving reference frames.

The primary trap is the failure to properly shift into the wedge's frame to apply the coefficient of restitution, and subsequently failing to shift back to the ground frame to find the true final velocity.

Students commonly calculate the relative post-collision velocity ( $20\hat{j}$ ) and incorrectly use it as the absolute velocity, yielding a max height of $100\text{ m}$ .

Another trap is assuming the collision doesn't alter the energy manifold, leading them to either use the initial projectile max height ( $80\text{ m}$ ) or assume an elastic collision ( $160\text{ m}$ ).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Maximum Height Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free