JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 20)
Two particles, of masses M and 2M, moving,
as shown, with speeds of 10 m/s and 5 m/s,
collide elastically at the origin. After the
collision, they move along the indicated
directions with speeds u1 and u2, respectively.
The values of u1 and u2 are nearly :
_10th_April_Morning_Slot_en_20_1.png)
6.5 m/s and 3.2 m/s
6.5 m/s and 6.3 m/s
3.2 m/s and 6.3 m/s
3.2 m/s and 12.6 m/s
Explanation
2MV, cos 30° + Mv2 cos 45° = 10 M cos 30° + 10 cos 45°
$$ \Rightarrow {v_1}\sqrt 3 + {{{v_2}} \over {\sqrt 2 }} = 5\sqrt 3 + 5\sqrt 2 $$ ...(i)
2MV, sin 30° – MV2 sin 45° = –10 M sin 30° + 10 M sin 45°
$${V_1} - {{{V_2}} \over {\sqrt 2 }} = - 5 + 5\sqrt 2 $$ ...(ii)
$${V_1} = {{5\left( {\sqrt 3 - 1} \right) + 10\sqrt 2 } \over {\sqrt 3 + 1}} = {{17.5} \over {2.7}} = 6.5\,m/s$$
V2 = 6.3 m/s
$$ \Rightarrow {v_1}\sqrt 3 + {{{v_2}} \over {\sqrt 2 }} = 5\sqrt 3 + 5\sqrt 2 $$ ...(i)
2MV, sin 30° – MV2 sin 45° = –10 M sin 30° + 10 M sin 45°
$${V_1} - {{{V_2}} \over {\sqrt 2 }} = - 5 + 5\sqrt 2 $$ ...(ii)
$${V_1} = {{5\left( {\sqrt 3 - 1} \right) + 10\sqrt 2 } \over {\sqrt 3 + 1}} = {{17.5} \over {2.7}} = 6.5\,m/s$$
V2 = 6.3 m/s
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