Question

A body of mass M moving at speed \(V_0\) collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles \(\theta_1\) and \(\theta_2\) with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles \(\theta_1\) and \(\theta_2\) will be equal, is :

• A. 2
• B. 3
• C. 1
• D. 4

Hint:

For equal masses (\(M=m\)) in an elastic collision with one at rest, the particles always move at right angles (\(\theta_1 + \theta_2 = 90^\circ\)).

Answer 1

The Correct Option is B

The problem involves a perfectly elastic collision between two masses. Let's solve it step by step:

1. When two bodies collide elastically, both momentum and kinetic energy are conserved. Consider the following for initial and final conditions:

• Initial momentum: MV_0
• Initial kinetic energy: \frac{1}{2}MV_0^2

2. After the collision, the momentum in terms of angles becomes:
   • Momentum conservation equations:
     M V_{0} = M V_1 \cos \theta_1 + m V_2 \cos \theta_2
   • In the perpendicular direction:
     0 = M V_1 \sin \theta_1 - m V_2 \sin \theta_2
3. Energy conservation:
   • \frac{1}{2}MV_0^2 = \frac{1}{2}MV_1^2 + \frac{1}{2}mV_2^2
4. If the angles are equal, \theta_1 = \theta_2 = \theta, the momentum equations can be solved:
   • Equate the sin terms for the perpendicular momentum:
     V_1 \sin \theta = \frac{m}{M} V_2 \sin \theta
   • Equate the cos terms for the parallel momentum:
     V_1 \cos \theta + \frac{m}{M} V_2 \cos \theta = V_0
5. To find the largest possible value of \frac{M}{m} where \theta_1 = \theta_2, simplify using the identity formulas:
   V_1 = V_2 = V', so:
   \frac{V'}{V_0} = \frac{\cos \theta}{\frac{M}{m} + 1}

Solving these equations gives, when maximizing the cosine function (cosine max is 1 when angle is zero, so mass must be comparable):

The angle condition yields:

• \frac{M}{m} = 3

7. Correct option: 3

Thus, the largest value of \frac{M}{m} for angles \theta_1 and \theta_2 to be equal is 3.