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A body of mass 6kg travelling with a vel...

A body of mass 6kg travelling with a velocity 10 m/s collides head - on and elastically with a body of mass 4kg travelling at a speed 5 m/s in opposite direction. The velocity of the second body after the collision is

A

0 m/s.

B

6 m/s

C

8 m/s

D

`13 ms^(-1)`

Text Solution

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The correct Answer is:
To solve the problem of the elastic collision between two bodies, we will use the principle of conservation of momentum and the formula for elastic collisions. Here’s the step-by-step solution: ### Step 1: Identify the given values - Mass of the first body (m1) = 6 kg - Initial velocity of the first body (u1) = 10 m/s - Mass of the second body (m2) = 4 kg - Initial velocity of the second body (u2) = -5 m/s (negative because it is in the opposite direction) ### Step 2: Write the formula for the final velocity of the second body (v2) after an elastic collision The formula for the final velocity of the second body after a head-on elastic collision is: \[ v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2} \] ### Step 3: Substitute the known values into the formula Substituting the values into the formula: \[ v_2 = \frac{(4 \, \text{kg} - 6 \, \text{kg})(-5 \, \text{m/s}) + 2 \cdot 6 \, \text{kg} \cdot 10 \, \text{m/s}}{6 \, \text{kg} + 4 \, \text{kg}} \] ### Step 4: Calculate the numerator and denominator Calculating the numerator: \[ = (4 - 6)(-5) + 2 \cdot 6 \cdot 10 \] \[ = (-2)(-5) + 120 \] \[ = 10 + 120 = 130 \] Calculating the denominator: \[ = 6 + 4 = 10 \] ### Step 5: Calculate v2 Now substituting back into the equation for v2: \[ v_2 = \frac{130}{10} = 13 \, \text{m/s} \] ### Final Answer The velocity of the second body after the collision is **13 m/s**. ---
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