Center of Mass , Momentum and Collision - NEET Physics Chapterwise MCQs & PYQs
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NEET Center of Mass , Momentum and Collision MCQs & PYQs
Practice NEET Center of Mass , Momentum and Collision Questions
Question 1:
moderate
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, \( 1 \text{ kg} \) first part moving with a velocity of \( 12 \text{ m s}^{-1} \) and \( 2 \text{ kg} \) second part moving with a velocity of \( 8 \text{ m s}^{-1} \). If the third part flies off with a velocity of \( 4 \text{ m s}^{-1} \), its mass would be:
(2009)
By conservation of momentum, \( P_{text{total}} = 0 \). Momentum of first part \( P_1 = 1 \text{ kg} \times 12 \text{ m/s} = 12 \text{ Ns} \). Momentum of second part \( P_2 = 2 \text{ kg} \times 8 \text{ m/s} = 16 \text{ Ns} \). As \( P_1 \) and \( P_2 \) are perpendicular, their resultant \( P_{12} = sqrt{12^2 + 16^2} = 20 \text{ Ns} \). For conservation, \( P_3 \) must be \( 20 \text{ Ns} \). \( m_3 = P_3 / v_3 = 20 \text{ Ns} / 4 \text{ m/s} = 5 \text{ kg} \).
A mass of \( 1 \text{ kg} \) is thrown up with a velocity of \( 100 \text{ m/s} \). After \( 5 \) seconds, it explodes into two parts. One part of mass \( 400 \text{ g} \) comes down with a velocity \( 25 \text{ m/s} \). Calculate the velocity of other part:
(2000)
Velocity of \( 1 \text{ kg} \) mass after \( 5 \text{ s} \): \( v = u - gt = 100 - 10 \times 5 = 50 \text{ m/s} \) (upward). Initial momentum before explosion \( P_i = 1 \text{ kg} \times 50 \text{ m/s} = 50 \text{ Ns} \) (upward). Mass of first part \( m_1 = 0.4 \text{ kg} \), \( v_1 = -25 \text{ m/s} \). Mass of second part \( m_2 = 0.6 \text{ kg} \). By conservation of momentum: \( P_i = m_1 v_1 + m_2 v_2 \). \( 50 = 0.4 \times (-25) + 0.6 v_2 \). \( 50 = -10 + 0.6 v_2 \implies v_2 = 100 \text{ m/s} \) (upward).
A body of mass \(4m\) is lying in \(x-y\) plane at rest. It suddenly explodes into three pieces. Two pieces each of mass \(m\) move perpendicular to each other with equal speeds \(v\). The total kinetic energy generated due to explosion is:
(2014)
Initial momentum is zero. Two pieces of mass \(m\) move with velocity \(v\) perpendicular to each other. Their momenta are \(m\vec{v}_1 = mv\hat{i}\, m\vec{v}_2 = mv\hat{j}\). The third piece has mass \(m_3 = 4m - m - m = 2m\). By momentum conservation, \(m_3\vec{v}_3 = -(mv\hat{i} + mv\hat{j})\), so \(|\vec{v}_3| = \frac{\sqrt{(mv)^2 + (mv)^2}}{2m} = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}\). Total KE = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass \(1\text{ kg}\) moves with a speed of \(12\text{ ms}^{-1}\) and the second part of mass \(2\text{ kg}\) moves with \(8\text{ ms}^{-1}\) speed. If the third part flies off with \(4\text{ ms}^{-1}\) speed, then its mass is:
(2013, 2009)
By conservation of momentum, the initial momentum is zero. Momentum of first part \(p_1 = 1\text{ kg} \times 12\text{ m/s} = 12\text{ kg m/s}\). Momentum of second part \(p_2 = 2\text{ kg} \times 8\text{ m/s} = 16\text{ kg m/s}\). Since they are perpendicular, resultant momentum \(p_{12} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ kg m/s}\). The third part must have momentum \(p_3 = 20\text{ kg m/s}\). Given its speed \(v_3 = 4\text{ m/s}\), its mass \(m_3 = p_3/v_3 = 20/4 = 5\text{ kg}\).
A shell of mass \(200\text{ gm}\) is ejected from a gun of mass \(4\text{ kg}\) by an explosion that generates \(1.05\text{ kJ}\) of energy. The initial velocity of the shell is
(2008)
Let shell mass \(m_s = 0.2\text{ kg}\), gun mass \(m_g = 4\text{ kg}\). Energy \(E = 1050\text{ J}\). By momentum conservation \(m_s v_s = m_g v_g\), so \(v_g = \frac{m_s v_s}{m_g} = \frac{0.2 v_s}{4} = \frac{v_s}{20}\). The energy is KE: \(E = \frac{1}{2}m_s v_s^2 + \frac{1}{2}m_g v_g^2 = \frac{1}{2}(0.2)v_s^2 + \frac{1}{2}(4)(\frac{v_s}{20})^2 = 0.1v_s^2 + \frac{2v_s^2}{400} = 0.1v_s^2 + 0.005v_s^2 = 0.105v_s^2\). Thus, \(v_s^2 = \frac{1050}{0.105} = 10000\), so \(v_s = 100\text{ m/s}\).
Body A of mass $4m$ moving with speed $u$ collides with another body B of mass $2m$, at rest. The collision is head-on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
(2019)
Final velocity of A is $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u = \frac{1}{3}u$. Fraction of energy lost is $1 - \left(\frac{v_1}{u}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
A moving block having mass $m$, collides with another stationary block having mass $4m$. The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$, then the value of coefficient of restitution ($e$) will be
A ball is dropped from a height of $5\text{ m}$, if it rebound upto height of $1.8\text{ m}$, then the ratio of velocities of the ball after and before rebound is:
(1998)
The velocity before rebound is $v_1 = \sqrt{2gh_1}$ and after rebound is $v_2 = \sqrt{2gh_2}$. The ratio is $\frac{v_2}{v_1} = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{1.8}{5}} = \frac{3}{5}$.
A metal ball of mass $2\text{ kg}$ moving with speed of $36\text{ km/h}$ has a head on collision with a stationary ball of mass $3\text{ kg}$. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is:
(1997)
Initial kinetic energy $K_i = \frac{1}{2}m_1 u_1^2 = 100\text{ J}$ (with $u_1 = 10\text{ m/s}$). Final velocity $v = \frac{m_1 u_1}{m_1+m_2} = 4\text{ m/s}$, and final kinetic energy $K_f = \frac{1}{2}(m_1+m_2)v^2 = 40\text{ J}$. Loss in K.E. = $100 - 40 = 60\text{ J}$.
A moving body of mass $m$ and velocity $3\text{ km/hour}$ collides with a rest body of mass $2\text{ m}$ and sticks to it. Now the combined mass starts to move. What will be the combined velocity?
(1996)
Using conservation of linear momentum, $mu_1 = (m+2m)v$, where $u_1 = 3\text{ km/hour}$. Solving gives $3m = 3mv \implies v = 1\text{ km/hour}$