In the post dated 10^{th} July 2007, a questions on two dimensional motion was given without solution. The question with solution is given below as promised in the post:

A steel sphere A projected up with a velocity of 10 ms^{–1} at an angle of 60º with the horizontal, collides elastically with an identical steel sphere B located at the highest point of the trajectory of A. The sphere B is the bob of a simple pendulum. After the collision, if the sphere B just moves along a vertical circle, what is the length of the pendulum? ( Take g = 10 ms^{–2})

(a) 50 cm (b) 55 cm (c) 60 cm (d) 65 cm (e) 70 cm

At the highest point of the trajectory, the velocity of the projectile is horizontal and is equal to **ucosθ **= 10 cos 60° = 5 ms^{–1}. Since the collision is elastic and the spheres are identical, the sphere A transfers its entire momentum to sphere B. Sphere B therefore moves with the *same* horizontal velocity of 5 ms^{–1}.

If the sphere is to just move along a vertical circle, the minimum velocity required at the lowest point of the circle is √(5gR) where ‘R’ is the radius of the circle (which is the length of the pendulum here). Therefore, we have

√(5gR) = 5 ms^{–1}.

On substituting for the acceleration due to gravity, g (=10 ms^{–1}), R = 0.5 m = 50 cm.

Now, consider the following MCQ:

**A ball rolling along a horizontal floor with a velocity of 7 ms**^{–1} reaches the edge of a long stair case from where it moves down as a projectile. If the width and the height of the steps of the stair case are 25 cm and 10 cm respectively, the ball will strike the edge of the n^{th} step from the top, where n is equal to

**(a) 10 (b) 12 (c) 14 (d) 16 (e) 18 **

Since the horizontal velocity is unchanged throughout the trajectory of the ball, we have

7 t = 0.25 n where ‘t’ is the time taken to hit the edge of the n^{th} step after leaving the top of the stair case.

Therefore, t = 0.25 n / 7

The vertical fall of the ball during this time is ½ g t^{2} and we have

½ g t^{2} = 0.1 n

Substituting for t and g, ½ ×9.8×0.0625 n^{2}/49 = 0.1n

Therefore, n = (49×0.1)/(4.9×0.0625) =16

One of my favorite problems is the bowling ball problem - you throw the ball horizontally, not rolling, at the ground with an initial velocity of v_0. It slides along the ground with a given coefficient of friction for an unknown distance and time, and then begins rolling. Find the time and distance to begin rolling in terms of things like μ, g, and v_0.

ReplyDeleteI like this one so much b/c it combines not only 2-D kinematics, but also rotational motion, forces, torque, friction, and just about everything. It's a good one to assess whether a student knows many major topics and can combine them. Even when you've seen and solved the problem before, it's still a challenging problem.

Your favorite problem will be fascinating to anybody genuinely interested in Physics. Let me see whether I can include such questions (qualitative ones at least) at physicsplus. The problem is that the questions here are expected to be worked out in a relatively short time as in the context of entrance examinations.

ReplyDeleteThank you so much.