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Collisions
Appears in
Concepts tested here
- Coefficient of restitution
- Energy conservation in inelastic collision
- Perfectly inelastic collision in two dimensions
All Questions
2015 AIPMT-I 1 question
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Two particles of masses $m_1$, $m_2$ move with initial velocities $u_1$ and $u_2$. On collision, one of the particles gets excited to a higher level, after absorbing energy $\varepsilon$. If final velocities of particles be $v_1$ and $v_2$ then we must haveTotal energy is conserved in any collision:
Initial KE = Final KE + excitation energy
$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \varepsilon$
$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$
2011 AIPMT-MAINS 1 question
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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination isThe masses stick together, so the collision is perfectly inelastic; linear momentum is conserved.
Initial momentum = $m(v\hat{i}) + 3m(2v\hat{j}) = mv\hat{i} + 6mv\hat{j}$
Final momentum = $(m + 3m)\vec{V} = 4m\vec{V}$
$4m\vec{V} = mv\hat{i} + 6mv\hat{j}$
$\vec{V} = \frac{mv\hat{i} + 6mv\hat{j}}{4m}$
$\vec{V} = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$
2010 AIPMT-PRE 1 question
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A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will beLet the masses be m and 2m and the velocities after collision be $v_1$ and $v_2$.
Conservation of momentum: $m\times2 + 2m\times0 = mv_1 + 2mv_2$, so $v_1 + 2v_2 = 2$ ... (i)
Coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$, so $v_2 - v_1 = 0.5\times(2 - 0) = 1$ ... (ii)
Adding (i) and (ii): $3v_2 = 3$, so $v_2 = 1$ m/s
From (ii): $v_1 = v_2 - 1 = 0$
The velocities after collision are 0 and 1 m/s.