© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the initial conditions and the final requirement
• A particle of mass m moves initially with speed u along the x-axis (its velocity is u\hat{i} ).
• It collides elastically with another particle of mass 10m that is initially at rest.
• After the collision, the smaller mass has half its initial kinetic energy.
• We denote the angles of the scattered particles (with respect to the x-axis) by \theta_1 (for the smaller mass) and \theta_2 (for the heavier mass).
• We need to find n if \sin\theta_1 = \sqrt{n}\,\sin\theta_2 .
Step 2: Express the final speeds using kinetic energy conservation
• Initial kinetic energy of the smaller mass:
K_{\text{initial}} = \tfrac{1}{2} m u^2.
• The small mass is left with half this initial kinetic energy after collision, so its final kinetic energy is
K_{1,\text{final}} = \tfrac{1}{2} \times \tfrac{1}{2} m u^2 = \tfrac{1}{4} m u^2.
• Hence, the final speed of the smaller mass ( v_1 ) is found by
\tfrac{1}{2} m v_1^2 = \tfrac{1}{4} m u^2
\quad\Longrightarrow\quad
v_1 = \tfrac{u}{\sqrt{2}}.
• Because the collision is elastic, total kinetic energy is conserved. Therefore, the heavier mass ( 10m ) must acquire the remaining kinetic energy:
K_{2,\text{final}} = K_{\text{initial}} - K_{1,\text{final}}
= \tfrac{1}{2} m u^2 - \tfrac{1}{4} m u^2
= \tfrac{1}{4} m u^2.
• Thus,
\tfrac{1}{2} \,(10m)\, v_2^2 = \tfrac{1}{4} m u^2
\quad\Longrightarrow\quad
5m\,v_2^2 = \tfrac{1}{4} m u^2
\quad\Longrightarrow\quad
v_2^2 = \tfrac{1}{20} u^2
\quad\Longrightarrow\quad
v_2 = \tfrac{u}{\sqrt{20}} = \tfrac{u}{2\sqrt{5}}.
Step 3: Apply conservation of momentum in the x- and y-directions
Let the final velocities be \vec{v}_1 (magnitude v_1 ) for the smaller mass and \vec{v}_2 (magnitude v_2 ) for the heavier mass, making angles \theta_1 and \theta_2 with the x-axis respectively.
1) In the x-direction:
m u = m\,v_1 \cos\theta_1 + (10m)\,v_2 \cos\theta_2.
Substituting v_1 = \tfrac{u}{\sqrt{2}} and v_2 = \tfrac{u}{2\sqrt{5}} ,
m u
= m \left(\tfrac{u}{\sqrt{2}}\right)\cos\theta_1 + 10m \left(\tfrac{u}{2\sqrt{5}}\right)\cos\theta_2.
Divide through by m u :
1
= \tfrac{1}{\sqrt{2}}\cos\theta_1 + \tfrac{10}{2\sqrt{5}} \cos\theta_2
= \tfrac{1}{\sqrt{2}}\cos\theta_1 + \sqrt{5}\,\cos\theta_2.
\tag{1}
2) In the y-direction (initial y-momentum is zero):
0 = m\,v_1 \sin\theta_1 + (10m)\,v_2 \sin\theta_2.
Substituting the values of v_1 and v_2 again:
0
= m \left(\tfrac{u}{\sqrt{2}}\right) \sin\theta_1
+ 10m \left(\tfrac{u}{2\sqrt{5}}\right) \sin\theta_2.
Divide through by m \tfrac{u}{\sqrt{2}} :
0
= \sin\theta_1
+ 10 \,\tfrac{\tfrac{u}{2\sqrt{5}}}{\tfrac{u}{\sqrt{2}}}\,\sin\theta_2
= \sin\theta_1
+ 10 \,\tfrac{1}{2\sqrt{5}} \tfrac{\sqrt{2}}{1}\,\sin\theta_2
= \sin\theta_1 + \sqrt{10}\,\sin\theta_2.
Hence,
\sin\theta_1 = -\,\sqrt{10}\,\sin\theta_2.
By magnitude consideration (often stated as \sin\theta_1 = \sqrt{10}\,\sin\theta_2 depending on direction), we identify
\sin\theta_1 = \sqrt{10}\,\sin\theta_2.
Step 4: Identify the value of n
We are given that
\sin\theta_1 = \sqrt{n}\,\sin\theta_2.
From the momentum conservation in the y-direction, we have
\sin\theta_1 = \sqrt{10}\,\sin\theta_2.
Therefore, by comparison, \sqrt{n} = \sqrt{10} , which implies
n = 10.
Final Answer
The value of n is 10.
