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

Step 1: Understanding the Concept:
In an elastic collision, both momentum and kinetic energy are conserved. If the scattering angles are equal (\(\theta_1 = \theta_2 = \theta\)), we can use conservation laws to find the constraints on the mass ratio.
Step 2: Key Formula or Approach:
Conservation of momentum (x and y directions):
\( MV_0 = MV_1 \cos\theta + mV_2 \cos\theta \implies MV_0 = (MV_1 + mV_2) \cos\theta \)
\( 0 = MV_1 \sin\theta - mV_2 \sin\theta \implies MV_1 = mV_2 \implies V_2 = \frac{M}{m}V_1 \)
Step 3: Detailed Explanation:
Substitute \( V_2 \) into the x-momentum equation:
\( MV_0 = (MV_1 + m \frac{M}{m} V_1) \cos\theta = 2MV_1 \cos\theta \implies V_1 = \frac{V_0}{2\cos\theta} \)
Conservation of Energy:
\( \frac{1}{2} MV_0^2 = \frac{1}{2} MV_1^2 + \frac{1}{2} mV_2^2 \)
\( MV_0^2 = M \left( \frac{V_0}{2\cos\theta} \right)^2 + m \left( \frac{M}{m} \frac{V_0}{2\cos\theta} \right)^2 \)
Divide by \( \frac{MV_0^2}{4\cos^2\theta} \):
\( 4\cos^2\theta = 1 + \frac{M}{m} \)
Since the maximum value of \(\cos^2\theta\) is 1:
\( 1 + \frac{M}{m} \leq 4 \times 1 \)
\( \frac{M}{m} \leq 3 \)
Step 4: Final Answer:
The largest possible value of the ratio M/m is 3.