Question

Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:

• A. \(\frac19\)
• B. \(\frac89\)
• C. \(\frac49\)
• D. \(\frac59\)

Answer 1

The Correct Option is B

 To solve this problem, we need to understand the concept of elastic collisions. In an elastic collision, both momentum and kinetic energy are conserved. Let's break down the problem step-by-step:

We have two bodies involved in the collision:

• Body A: Mass \(4m\) and initial velocity \(u\).
• Body B: Mass \(2m\) and initially at rest (velocity = 0).

Since the collision is elastic, both momentum and kinetic energy are conserved. First, let's write the equation for the conservation of momentum:

1. \(4m \cdot u = 4m \cdot v_1 + 2m \cdot v_2\)

Where \(v_1\) and \(v_2\) are the velocities of bodies A and B after the collision, respectively.

Now, let's write the equation for the conservation of kinetic energy:

1. \(\frac{1}{2} \cdot 4m \cdot u^2 = \frac{1}{2} \cdot 4m \cdot v_1^2 + \frac{1}{2} \cdot 2m \cdot v_2^2\)

Using these two conservation equations, we solve for \(v_1\) and \(v_2\). After simplifying, we find:

1. \(v_1 = \frac{u}{3}, \quad v_2 = \frac{4u}{3}\)

Next, we calculate the initial kinetic energy of body A:

1. \(KE_{\text{initial}} = \frac{1}{2} \cdot 4m \cdot u^2 = 2mu^2\)

The kinetic energy of body A after the collision is:

1. \(KE_{\text{final}} = \frac{1}{2} \cdot 4m \cdot \left( \frac{u}{3} \right)^2 = \frac{2mu^2}{9}\)

The energy lost by body A is:

1. \(\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 2mu^2 - \frac{2mu^2}{9} = \frac{18mu^2}{9} - \frac{2mu^2}{9} = \frac{16mu^2}{9}\)

The fraction of energy lost by body A is:

1. \(\frac{\Delta KE}{KE_{\text{initial}}} = \frac{\frac{16mu^2}{9}}{2mu^2} = \frac{16}{18} = \frac{8}{9}\)

Therefore, the fraction of energy lost by the colliding body A is \(\frac{8}{9}\). The correct answer is

\(\frac{8}{9}\)

.