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Question

Physics Question on work, energy and power

Two spheres AA and BB of masses m1m_1 and m2m_2 respectively collide. AA is at rest initially and BB is moving with velocity v/2v/2 along x-axis. After collision BB has a velocity in a direction perpendicular to the original direction. The mass AA moves after collision in the direction

A

same as that of B

B

opposite to that of B

C

θ=tan1(12)\theta=tan^{-1}\bigg(\frac{1}{2}\bigg) to the x-axis

D

θ=tan1(12)\theta=tan^{-1}\bigg(-\frac{1}{2}\bigg) to the x-axis

Answer

θ=tan1(12)\theta=tan^{-1}\bigg(-\frac{1}{2}\bigg) to the x-axis

Explanation

Solution

There is no external force acting on the spheres. So linear momentum will be conserved.
Before the collision, In direction xx, Linear momentum =m2v= m _{2} v....1
In direction yy, Linear momentum =0=0
After the collision, spheres moves as shown in figure. Let velocity of sphere A is
v1v _{1}
In direction xx, Linear momentum =m1v1cos(θ)2= m _{1} v _{1} \cos (\theta) \ldots 2
In direction y, Linear momentum =m2v2m1v1sin(θ).3=\frac{ m _{2} v }{2}- m _{1} v _{1} \sin (\theta) \ldots \ldots .3
Linear momentum will be conserved,
From equation 1 and 2,m1v1cos(θ)=m2v42, \Rightarrow m_{1} v_{1} \cos (\theta)=m_{2} v \ldots 4
From the equation 2, m2v2m1v1sin(θ)=0m1v1sin(θ)=m2v25\Rightarrow \frac{ m _{2} v }{2}- m _{1} v _{1} \sin (\theta)=0 \Rightarrow m _{1} v _{1} \sin (\theta)=\frac{ m _{2} v }{2} \ldots \ldots 5
Dividing equation 5 by 4,tan(θ)=12θ=tan1(12)4, \Rightarrow \tan (\theta)=\frac{1}{2} \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)