JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 6)
A body of mass M moving at speed V0 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 :
4
1
3
2
Explanation
_31st_August_Morning_Shift_en_6_1.png)
Given $$\theta$$1 = $$\theta$$2 = $$\theta$$
from momentum conservation
in x-direction MV0 = MV1 cos$$\theta$$ + mV2 cos$$\theta$$
in y-direction 0 = MV1 sin$$\theta$$ $$-$$ mV2 sin$$\theta$$
Solving above equations
$${V_2} = {{M{V_1}} \over m}$$, V0 = 2V1 cos$$\theta$$
From energy conservation
$${1 \over 2}MV_0^2 = {1 \over 2}MV_1^2 + {1 \over 2}MV_2^2$$
Substituting value of V2 & V0, we will get
$${M \over m} + 1 = 4{\cos ^2}\theta \le 4$$
$${M \over m} \le 3$$
Option (c)
Comments (0)
