JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Conservation of Momentum

An alpha-particle of mass

m

suffers 1-deminsional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is :

(A)2 m
(B)3.5 m
(C)1.5 m
(D)4 m✓ Correct

Answer: (D)

Solution

In a head-on elastic collision the projectile's velocity ratio is

\frac{v _{f}}{v _{i}} = \frac{m - M}{m + M}

. Losing

64%

of its kinetic energy leaves

36%

, so

(\frac{m - M}{m + M})^{2} = 0.36

. Since the

α

-particle rebounds straight back, take the negative root,

\frac{m - M}{m + M} = - \frac{3}{5}

. Then

5 (m - M) = - 3 (m + M)

, i.e.

5 m - 5 M = - 3 m - 3 M

, giving

8 m = 2 M

and

M = 4 m

. The nucleus is four times as massive as the

α

-particle.

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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