Question

If no external force is acting on the system, net linear momentum of the system is conserved. If system is acted upon by some external force, the component of momentum of the system, along which no external force is present or their vector sum is zero, is conserved. If a sharp blow is given to a body its linear momentum changes immediately. Change in angular momentum not only depends on the magnitude of the blow but also on point of application. In the case of symmetrical body we take the axis of rotation through center of the body. A wedge of mass $4 m$ is placed at rest on a smooth horizontal surface. A uniform solid sphere of mass $m$ and radius $r$ is placed at rest on the flat portion of the wedge at the point $Q$ as shown in the figure. A sharp horizontal impulse $P$ is given to the sphere at a point below $h=0.4 r$ from the center of the sphere. The radius of curvature of the curved portion of the wedge is $R$. Coefficient of friction to the left side of point $Q$ is $\mu$ and to the right side of point $Q$ is zero. For a body to roll on a surface without slipping, there should be no relative velocity between the points of contact.

The maximum height to which the center of mass of the sphere will climb on the curved portion of the wedge is :

• A. $\frac{2 P^{2}}{5 m^{2} g}$
• B. $\frac{P^{2}}{5 m^{2} g}$
• C. $\frac{P^{2}}{2 m^{2} g}$
• D. none of these

Answer 1

Answer: (A)

All maximum height when system is moving at speed $v_{1}$ by conservation of energy
$P=5 m v_{1}=m v$
$\frac{1}{2} m v^{2}=\frac{1}{2}(5 m) v_{1}^{2}+m g h$
$\Rightarrow \frac{P^{2}}{2 m}=\frac{P^{2}}{10 m}+m g h .$
$\Rightarrow h=\frac{4 P^{2}}{10 m^{2} g}=\frac{2 P^{2}}{5 m^{2} g}$