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Body A A A of mass 4 m 4m 4 m moving with speed u u u collides with another body B B B of mass 2 m 2m 2 m , at rest. The collision is head-on and elastic in nature. After the collision the fraction of energy lost by the colliding body A A A is:
medium
Work, Energy and Power
2019
physics
Explanation To solve this problem, we need to analyze the elastic collision between two bodies. \\
Given: \\
• Mass of body A = 4 m A = 4m \\
A = 4 m • Initial velocity of body A = u A = u \\
A = u • Mass of body B = 2 m B = 2m \\
B = 2 m
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• Initial velocity of body
• The collision is elastic.
In an elastic collision, both momentum and kinetic energy are conserved.
1. Conservation of momentum:
4 m ⋅ u + 2 m ⋅ 0 = 4 m ⋅ v A + 2 m ⋅ v B \\
4m \cdot u + 2m \cdot 0 = 4m \cdot v_A + 2m \cdot v_B \\
4 m ⋅ u + 2 m ⋅ 0 = 4 m ⋅ v A + 2 m ⋅ v B Simplifying, we get:
4 u = 4 v A + 2 v B \\
4u = 4v_A + 2v_B \quad 4 u = 4 v A + 2 v B (Equation 1)
2. Conservation of kinetic energy:
1 2 ⋅ 4 m ⋅ u 2 = 1 2 ⋅ 4 m ⋅ v A 2 + 1 2 ⋅ 2 m ⋅ v B 2 \\
\frac{1}{2} \cdot 4m \cdot u^2 = \frac{1}{2} \cdot 4m \cdot v_A^2 + \frac{1}{2} \cdot 2m \cdot v_B^2 \\
2 1 ⋅ 4 m ⋅ u 2 = 2 1 ⋅ 4 m ⋅ v A 2 + 2 1 ⋅ 2 m ⋅ v B 2 Simplifying, we get:
4 u 2 = 4 v A 2 + 2 v B 2 \\
4u^2 = 4v_A^2 + 2v_B^2 \quad 4 u 2 = 4 v A 2 + 2 v B 2 (Equation 2)
From Equation 1, solve for
v B : v B = 2 u − 2 v A v_B: \\
v_B = 2u - 2v_A \\
v B : v B = 2 u − 2 v A Substitute
in Equation 2:
4 u 2 = 4 v A 2 + 2 ( 2 u − 2 v A ) 2 \\
4u^2 = 4v_A^2 + 2(2u - 2v_A)^2 \\
4 u 2 = 4 v A 2 + 2 ( 2 u − 2 v A ) 2 Simplify:
4 u 2 = 4 v A 2 + 2 ( 4 u 2 − 8 u v A + 4 v A 2 ) 4 u 2 = 4 v A 2 + 8 u 2 − 16 u v A + 8 v A 2 4 u 2 = 12 v A 2 − 16 u v A + 8 u 2 \\
4u^2 = 4v_A^2 + 2(4u^2 - 8uv_A + 4v_A^2) \\
4u^2 = 4v_A^2 + 8u^2 - 16uv_A + 8v_A^2 \\
4u^2 = 12v_A^2 - 16uv_A + 8u^2 \\
4 u 2 = 4 v A 2 + 2 ( 4 u 2 − 8 u v A + 4 v A 2 ) 4 u 2 = 4 v A 2 + 8 u 2 − 16 u v A + 8 v A 2 4 u 2 = 12 v A 2 − 16 u v A + 8 u 2 Rearrange terms:
0 = 12 v A 2 − 16 u v A + 4 u 2 \\
0 = 12v_A^2 - 16uv_A + 4u^2 \\
0 = 12 v A 2 − 16 u v A + 4 u 2 Divide the entire equation by 4:
0 = 3 v A 2 − 4 u v A + u 2 \\
0 = 3v_A^2 - 4uv_A + u^2 \\
0 = 3 v A 2 − 4 u v A + u 2 This is a quadratic equation in
Solve using the quadratic formula:
v A = − b ± b 2 − 4 a c 2 a \\
v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
v A = 2 a − b ± b 2 − 4 a c Here,
a = 3 , b = − 4 u , c = u 2 . v A = 4 u ± ( − 4 u ) 2 − 4 ⋅ 3 ⋅ u 2 6 v A = 4 u ± 16 u 2 − 12 u 2 6 v A = 4 u ± 4 u 2 6 v A = 4 u ± 2 u 6 a = 3, \, b = -4u, \, c = u^2. \\
v_A = \frac{4u \pm \sqrt{(-4u)^2 - 4 \cdot 3 \cdot u^2}}{6} \\
v_A = \frac{4u \pm \sqrt{16u^2 - 12u^2}}{6} \\
v_A = \frac{4u \pm \sqrt{4u^2}}{6} \\
v_A = \frac{4u \pm 2u}{6} \\
a = 3 , b = − 4 u , c = u 2 . v A = 6 4 u ± ( − 4 u ) 2 − 4 ⋅ 3 ⋅ u 2 v A = 6 4 u ± 16 u 2 − 12 u 2 v A = 6 4 u ± 4 u 2 v A = 6 4 u ± 2 u This gives two solutions:
v A = 6 u 6 = u \\
v_A = \frac{6u}{6} = u \quad v A = 6 6 u = u (not possible as
must lose some speed)
v A = 2 u 6 = u 3 \\
v_A = \frac{2u}{6} = \frac{u}{3} \\
v A = 6 2 u = 3 u Thus,
v A = u 3 . v_A = \frac{u}{3}. \\
v A = 3 u . Now, calculate the initial and final kinetic energy of body
Initial kinetic energy of
A = 1 2 ⋅ 4 m ⋅ u 2 = 2 m u 2 A = \frac{1}{2} \cdot 4m \cdot u^2 = 2mu^2 \\
A = 2 1 ⋅ 4 m ⋅ u 2 = 2 m u 2 Final kinetic energy of
A = 1 2 ⋅ 4 m ⋅ ( u 3 ) 2 = 2 m u 2 9 A = \frac{1}{2} \cdot 4m \cdot \left(\frac{u}{3}\right)^2 = \frac{2mu^2}{9} \\
A = 2 1 ⋅ 4 m ⋅ ( 3 u ) 2 = 9 2 m u 2 Energy lost by
A = 2 m u 2 − 2 m u 2 9 = 18 m u 2 9 − 2 m u 2 9 = 16 m u 2 9 A = 2mu^2 - \frac{2mu^2}{9} = \frac{18mu^2}{9} - \frac{2mu^2}{9} = \frac{16mu^2}{9} \\
A = 2 m u 2 − 9 2 m u 2 = 9 18 m u 2 − 9 2 m u 2 = 9 16 m u 2 Fraction of energy lost by
A = E n e r g y l o s t I n i t i a l e n e r g y = 16 m u 2 9 2 m u 2 = 8 9 A = \frac{Energy \ lost}{Initial \ energy} = \frac{\frac{16mu^2}{9}}{2mu^2} = \frac{8}{9} \\
A = I ni t ia l e n er g y E n er g y l os t = 2 m u 2 9 16 m u 2 = 9 8 Therefore, the fraction of energy lost by body
is
This corresponds to Option 2.