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Question: A sphere p of mass m and velocity \(v_{1}\)undergoes an oblique and perfectly elastic collision with...

A sphere p of mass m and velocity v1v_{1}undergoes an oblique and perfectly elastic collision with an identical sphere Q initially at rest. The angle θ\thetabetween the velocities of the spheres after the collision shall be:

A

0

B

45045^{0}

C

1800180^{0}

D

90090^{0}

Answer

1800180^{0}

Explanation

Solution

According to law of conservation of linear momentum we get

mv→1+m×0=mv→pf+mv→Qfm{\overrightarrow{v}}_{1} + m \times 0 = m{\overrightarrow{v}}_{pf} + m{\overrightarrow{v}}_{Qf}

Where v→pfandv→Qf{\overrightarrow{v}}_{pf}and{\overrightarrow{v}}_{Qf}are the final velocities of spheres p and Q after collision respectively.

v→i=v→Pf+v→Qf{\overrightarrow{v}}_{i} = {\overrightarrow{v}}_{Pf} + {\overrightarrow{v}}_{Qf}

(v→i.v→i)=(v→pf+v→Qf).(v→pf+v→Qf)({\overrightarrow{v}}_{i}.{\overrightarrow{v}}_{i}) = ({\overrightarrow{v}}_{pf} + {\overrightarrow{v}}_{Qf}).({\overrightarrow{v}}_{pf} + {\overrightarrow{v}}_{Qf})

=v→pf.v→pf+v→Qf.v→Qf+2v→pf.v→Qf= {\overrightarrow{v}}_{pf}.{\overrightarrow{v}}_{pf} + {\overrightarrow{v}}_{Qf}.{\overrightarrow{v}}_{Qf} + 2{\overrightarrow{v}}_{pf}.{\overrightarrow{v}}_{Qf}

Or vi2=vpf2+vQf2+2vpfvQfcos⁡θv_{i}^{2} = v_{pf}^{2} + v_{Qf}^{2} + 2v_{pf}v_{Qf}\cos\theta ….(i)

According to conservation of kinetic energy, we get

12mvi2=12mvpf2+12mvQf2\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{pf}^{2} + \frac{1}{2}mv_{Qf}^{2} …(ii)

vi2=vpf2+vQf2v_{i}^{2} = v_{pf}^{2} + v_{Qf}^{2}

Comparing (i) and (ii) we get

2vpfvlqfcos⁡θ=02v_{pf}vlqf\cos\theta = 0

⇒cos⁡θ=0orθ=900\Rightarrow \cos\theta = 0or\theta = 90^{0}