Assertion (A): In case of bullet fired from a gun, the ratio of kinetic energy of gun and bullet is equal to ratio of masses of bullet and gun.
Reason (R): In firing of bullet, linear momentum of system is conserved.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Reason (R): For the bullet-gun system, the forces causing the bullet to fire are internal. Thus, linear momentum of the system is conserved. So, (R) is true.
Assertion (A): Let (m\) and (M\) be masses of bullet and gun, (v\) and (V\) their velocities. By momentum conservation, (mv = MV\). The ratio of kinetic energies is \( \frac{K_g}{K_b} = \frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2} = \frac{M(mv/M)^2}{mv^2} = \frac{m}{M}\). So, (A) is true.
(R) correctly explains (A) as the kinetic energy ratio is derived directly from momentum conservation. Option (1) is correct.
Consider the given statements and choose the correct option that follows:
Statement 1: During a collision the total linear momentum of system is conserved at each instant of collision.
Statement 2: During a collision the kinetic energy conservation holds always.
Based on above information, pick the correct option.
1. Both statements (1) and (2) are true
2. Both statements (1) and (2) are false
3. Statement (1) is true but (2) is false
4. Statement (1) is false but (2) is true
View Answer
Total linear momentum is conserved at each instant of collision because no external forces act. Kinetic energy, however, is not conserved during the period of deformation, and is conserved after only in perfectly elastic collisions. Thus, Statement 1 is true and Statement 2 is false.
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)
1. \(mv^2\)
2. \(3/2 mv^2\)
3. \(2 mv^2\)
4. \(4 mv^2\)
View Answer
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 = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).
A mass $m$ moving horizontally (along the $x$-axis) with velocity $v$ collides and sticks to a mass of $3\text{ m}$ moving vertically upward (along the $y$-axis) with velocity $2v$. The final velocity of the combination is:
(2011 Mains)
1. $\frac{3}{2}\hat{i} + \frac{1}{4}\hat{j}$
2. $\frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$
3. $\frac{1}{3}v\hat{i} + \frac{2}{3}v\hat{j}$
4. $\frac{2}{3}v\hat{i} + \frac{1}{3}v\hat{j}$
View Answer
By conservation of momentum, total initial momentum vector is $\vec{P} = mv\hat{i} + (3m)(2v)\hat{j}$. Dividing by the total mass $4m$ gives the final velocity vector $\vec{v}_f = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$.
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)
1. $0, 2$
2. $0, 1$
3. $1, 1$
4. $1, 0.5$
View Answer
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 $v_2 = 1\text{ m/s}$.