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 51:
moderate
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 moving with velocity $2\text{ m/s}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$ then their velocities (in $\text{ m/s}$) after collision will be:
(2010 Pre)
Using the collision velocity formulas $v_1 = \frac{(m_1 - em_2)u_1}{m_1+m_2}$ and $v_2 = \frac{(1+e)m_1 u_1}{m_1+m_2}$ with $m_1=m$, $m_2=2m$, $u_1=2$, and $e=0.5$, we get $v_1 = 0$ and $v_2 = 1\text{ m/s}$.
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 $3m = 3mv \implies v = 1\text{ km/hour}$
Two identical balls $A$ and $B$ having velocities of $0.5\text{ m/s}$ and $-0.3\text{ m/s}$ respectively collide elastically in one dimension. The velocities of $B$ and $A$ after the collision respectively will be:
(2016, 1998, 1994, 1991)
In an elastic collision between two identical bodies, their velocities are mutually exchanged. Given initial velocities are $u_1 = 0.5\text{ m/s}$ and $u_2 = -0.3\text{ m/s}$. Therefore, after collision, the velocities of $B$ and $A$ become $0.5\text{ m/s}$ and $-0.3\text{ m/s}$ respectively.