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IV. Numerical Problems · Q4

Q.Two different unknown masses AA and BB collide. AA is initially at rest when BB has a speed vv. After the collision BB has a speed v/2v/2 and moves at right angles to its original direction of motion. Find the direction in which AA moves after the collision.

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Step 1. Set up axes with BB's original direction of motion along the xx-axis. Before collision: AA is at rest (momentum =0=0); BB has momentum mBvm_Bv along +x+x.

Step 2. After collision, BB moves with speed v/2v/2 perpendicular to its original direction -- take this as the +y+y direction, so BB's final momentum is mB(v/2)j^m_B(v/2)\hat j, with zero xx-component remaining in BB.

Step 3. Momentum conservation along xx: mBv=mAvA,x+0⇒vA,x=mBmAvm_Bv = m_Av_{A,x} + 0 \Rightarrow v_{A,x}=\dfrac{m_B}{m_A}v.

Step 4. Momentum conservation along yy: 0=mAvA,y+mB(v/2)⇒vA,y=−mB2mAv0 = m_Av_{A,y} + m_B(v/2) \Rightarrow v_{A,y}=-\dfrac{m_B}{2m_A}v (negative, i.e. opposite to BB's deflection).

Step 5. The angle AA makes with the xx-axis (B's original direction): tan⁡θ=∣vA,yvA,x∣=(mB/2mA)v(mB/mA)v=12\tan\theta = \left|\dfrac{v_{A,y}}{v_{A,x}}\right| = \dfrac{(m_B/2m_A)v}{(m_B/m_A)v} = \dfrac{1}{2} -- notice the unknown mass ratio mB/mAm_B/m_A cancels out completely, so the angle is fixed regardless of what the (unknown) masses actually are. …

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