In FIGURE P11.57, a block of mass m slides along a frictionless track with speed ${\mathrm{v}}_{\mathrm{m}}$. It collides with a stationary block of mass M. Find an expression for the minimum value of ${\mathrm{v}}_{\mathrm{m}}$that will allow the second block to circle the loop-the-loop without falling off if the collision is (a) perfectly inelastic or (b) perfectly elastic.

We have given,

mass of moving block = m

Mass of big block = M

velocity of mass m =${\mathrm{v}}_{\mathrm{m}}$

We have to find the minimum velocity of mass m for which the bigger mass will take one circular motion along the loop when the collision is elastic.

Since the collision is given perfectly inelastic and one mass is at rest then after the Collison both the masses will move same.

The block will move along the loop if it has enough kinetic energy which should be equal to the potential energy.

When the mass will move along the loop it will experience the centripetal force such that it will balance the gravitational force.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{c}}={\mathrm{F}}_{\mathrm{G}} \\ \frac{(\mathrm{m}+\mathrm{M}){\mathrm{v}}^2}{\mathrm{R}}=(\mathrm{m}+\mathrm{M})\mathrm{g} \\ \mathrm{v}=\sqrt{\mathrm{gR}}.......\left(1\right) \end{gathered}$$

then,

By applying the conservation of energy at the top of the loop is such that ,

$$\begin{gathered} \frac{1}{2}(m+M)v^2=(m+M)gh+\frac{1}{2}(M+M)gR \\ v=\sqrt{5gR} \end{gathered}$$

Then,

Using the conservation of momentum,

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{m}}=(\mathrm{m}+\mathrm{M})\mathrm{v} \\ \mathrm{v}=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}}\sqrt{5\mathrm{gR}} \end{gathered}$$

We have given,

mass of moving block = m

Mass of big block = M

velocity of mass m =${\mathrm{v}}_{\mathrm{m}}$

We have to find the minimum velocity of mass m for which the bigger mass will take one circular motion along the loop when the collision is elastic.

Since the collision is given perfectly elastic and one mass is at rest then after the Collison another masses will move with some velocity v.

The block will move along the loop if it has enough kinetic energy which should be equal to the potential energy.

When the mass will move along the loop it will experience the centripetal force such that it will balance the gravitational force.

Then, we using the conservation of momentum,

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{m}}=\mathrm{Mv} \\ \mathrm{v}=\frac{{\mathrm{mv}}_{\mathrm{m}}}{\mathrm{M}} \end{gathered}$$

Now using the conservation of energy,

$$\begin{gathered} \frac{1}{2}m{v}_m^2=\frac{1}{2}M{v}^2 \\ {v}_m=\sqrt{\frac{M}{m}}v \end{gathered}$$

After the collision, By applying the conservation of energy at the top of the loop is such that ,

$$\begin{gathered} \frac{1}{2}(m+M)v^2=(m+M)gh+\frac{1}{2}(M+M)gR \\ v=\sqrt{5gR} \end{gathered}$$

then,

${\mathrm{v}}_{\mathrm{m}}=\frac{\mathrm{m}+\mathrm{M}}{2\mathrm{m}}\sqrt{5\mathrm{gR}}$