Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed \upsilon towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically 5 \over 6 th of the initial kinetic energy is lost in whole process. What is value of M/m ?

Center of Mass PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q53 with Solution

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Correct answer: C

So initial momentum is $p_i = mv$ and initial kinetic energy is $K_i = \frac12 mv^2.$

Since the collision is perfectly inelastic, $A$ and $B$ stick together.

Let their common speed after collision be $v_1$ .

Using conservation of momentum: $mv = (m+m)v_1 = 2m v_1$ $v_1 = \frac{v}{2}.$

So after first collision, kinetic energy is

= m\cdot \frac{v^2}{4} = \frac14 mv^2.$$ --- 3. **Second collision: combined mass $2m$ collides with $C$ perfectly inelastically** Now mass $2m$ moving with speed $v/2$ collides with block $C$ of mass $M$ at rest, and they stick together. Let final speed be $v_2$. Again, conserve momentum: $$(2m)\left(\frac{v}{2}\right) = (2m+M)v_2$$ $$mv = (2m+M)v_2$$ $$v_2 = \frac{mv}{2m+M}.$$ Hence final kinetic energy is $$K_f = \frac12 (2m+M)\left(\frac{mv}{2m+M}\right)^2 = \frac12 \cdot \frac{m^2v^2}{2m+M}.$$ --- 4. **Use given energy loss condition** Given that $\frac{5}{6}$ of the initial kinetic energy is lost in the whole process. So final kinetic energy left is $$K_f = \left(1-\frac56\right)K_i = \frac16 K_i.$$ Since $$K_i = \frac12 mv^2,$$ we get $$K_f = \frac16 \cdot \frac12 mv^2 = \frac{1}{12}mv^2.$$ Now equate this with expression found above: $$\frac12 \cdot \frac{m^2v^2}{2m+M} = \frac{1}{12}mv^2.$$ Cancel $v^2$: $$\frac12 \cdot \frac{m^2}{2m+M} = \frac{1}{12}m.$$ Multiply by $12$: $$6\cdot \frac{m^2}{2m+M} = m.$$ Divide by $m$: $$\frac{6m}{2m+M} = 1.$$ So, $$2m+M = 6m$$ $$M = 4m.$$ Therefore, $$\frac{M}{m} = 4.$$ --- 5. **Check options** - A: $5$ ✗ - B: $2$ ✗ - C: $4$ ✓ - D: $3$ ✗ So the correct option is **C**.

View written solutionFree

Correct answer: C

So initial momentum is $p_i = mv$ and initial kinetic energy is $K_i = \frac12 mv^2.$

Since the collision is perfectly inelastic, $A$ and $B$ stick together.

Let their common speed after collision be $v_1$ .

Using conservation of momentum: $mv = (m+m)v_1 = 2m v_1$ $v_1 = \frac{v}{2}.$

So after first collision, kinetic energy is

= m\cdot \frac{v^2}{4} = \frac14 mv^2.$$ --- 3. **Second collision: combined mass $2m$ collides with $C$ perfectly inelastically** Now mass $2m$ moving with speed $v/2$ collides with block $C$ of mass $M$ at rest, and they stick together. Let final speed be $v_2$. Again, conserve momentum: $$(2m)\left(\frac{v}{2}\right) = (2m+M)v_2$$ $$mv = (2m+M)v_2$$ $$v_2 = \frac{mv}{2m+M}.$$ Hence final kinetic energy is $$K_f = \frac12 (2m+M)\left(\frac{mv}{2m+M}\right)^2 = \frac12 \cdot \frac{m^2v^2}{2m+M}.$$ --- 4. **Use given energy loss condition** Given that $\frac{5}{6}$ of the initial kinetic energy is lost in the whole process. So final kinetic energy left is $$K_f = \left(1-\frac56\right)K_i = \frac16 K_i.$$ Since $$K_i = \frac12 mv^2,$$ we get $$K_f = \frac16 \cdot \frac12 mv^2 = \frac{1}{12}mv^2.$$ Now equate this with expression found above: $$\frac12 \cdot \frac{m^2v^2}{2m+M} = \frac{1}{12}mv^2.$$ Cancel $v^2$: $$\frac12 \cdot \frac{m^2}{2m+M} = \frac{1}{12}m.$$ Multiply by $12$: $$6\cdot \frac{m^2}{2m+M} = m.$$ Divide by $m$: $$\frac{6m}{2m+M} = 1.$$ So, $$2m+M = 6m$$ $$M = 4m.$$ Therefore, $$\frac{M}{m} = 4.$$ --- 5. **Check options** - A: $5$ ✗ - B: $2$ ✗ - C: $4$ ✓ - D: $3$ ✗ So the correct option is **C**.