Question 113: Centre of Mass and Rotational Motion

Question

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 $$	heta$$1 and $$	heta$$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 $$	heta$$1 and $$	heta$$2 will be equal, is :

Accepted Answer

Given $$	heta$$
1
= $$	heta$$
2
= $$	heta$$
from momentum conservation
in x-direction MV
0
= MV
1
cos$$	heta$$ + mV
2
cos$$	heta$$
in y-direction 0 = MV
1
sin$$	heta$$ $$-$$ mV
2
sin$$	heta$$
Solving above equations
$${V_2} = {{M{V_1}} \over m}$$, V
0
= 2V
1
cos$$	heta$$
From energy conservation
$${1 \over 2}MV_0^2 = {1 \over 2}MV_1^2 + {1 \over 2}MV_2^2$$
Substituting value of V
2
& V
0
, we will get
$${M \over m} + 1 = 4{\cos ^2}	heta  \le 4$$
$${M \over m} \le 3$$
Option (c)

Answer

4

Answer

1

Answer

3

Answer

2

Centre of Mass and Rotational Motion

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