Question 43: Centre of Mass and Rotational Motion

Question

In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg , moving in opposite directions with speeds of $10 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively, strike each other and stick together. The rise in temperature (in ${ }^{\circ} \mathrm{C}$ ), if all the heat produced during the collision is retained by these spheres, is :
(specific heat of sphere material $31 \mathrm{cal} / \mathrm{kg} .{ }^{\circ} \mathrm{C}$ and $1 \mathrm{cal}=4.2 \mathrm{~J}$ )

Accepted Answer

Before Collision :
The masses of the spheres are, $\mathrm{m}_{\mathrm{A}}=15 \mathrm{~kg}, \mathrm{~m}_2=25 \mathrm{~kg}$
The initial velocities of the spheres are, $\mathrm{u}_{\mathrm{A}}=10 \mathrm{~m} / \mathrm{s}, \mathrm{u}_{\mathrm{B}}=-30 \mathrm{~m} / \mathrm{s}$
Hence, the kinetic energy of system before collision is,
$$ \mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}^2+\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}^2 $$
$$\Rightarrow $$ $$ \mathrm{K}_{\mathrm{i}}=\frac{1}{2}(15)(10)^2+\frac{1}{2}(25)(30)^2=750+11250=12000 \mathrm{~J} $$
Since the collision is perfectly inelastic, the spheres stick together and move with a common final velocity $\mathrm{v}_{\mathrm{f}}$.
Using conservation of linear momentum,
$$ m_A u_A+m_B u_B=\left(m_A+m_Bight) v_f $$
$$\Rightarrow $$ $(15)(10)+(25)(-30)=(15+25) v_f$
$$\Rightarrow $$ $150-750=40 v_f$
$$ -600=40 v_f \Rightarrow v_f=-15 \mathrm{~m} / \mathrm{s} $$
(The negative sign indicates the combined mass moves in the direction of the 25 kg sphere.
So, the final kinetic energy of the system is
$$ \mathrm{K}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}ight) \mathrm{v}_{\mathrm{f}}^2 $$
$$\Rightarrow $$ $\mathrm{K}_{\mathrm{f}}=\frac{1}{2}(40)(15)^2$
$$\Rightarrow $$ $$ \mathrm{K}_{\mathrm{f}}=4500 \mathrm{~J} $$
The loss in kinetic energy appears as heat (Q).
$$ Q=K_i-K_f $$
$$\Rightarrow $$ $Q=12000 \mathrm{~J}-4500 \mathrm{~J}$
$$ Q=7500 \mathrm{~J} $$
Using the heat capacity formula $\mathrm{Q}=\operatorname{mc} \Delta \mathrm{T}$.
Where, total mass (m) of the spheres is 40 kg , the specific heat (c) of the material of spheres is $31 \mathrm{cal} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$.
As $1 \mathrm{cal}=4.2 \mathrm{~J}$, so $\mathrm{c}=31 	imes 4.2 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}=130.2 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$
Substituting the values in heat capacity formula,
$$ 7500=40 	imes 130.2 	imes \Delta \mathrm{T} $$
$$\Rightarrow $$ $7500=5208 	imes \Delta \mathrm{T}$
$$\Rightarrow $$ $$ \Delta \mathrm{T}=\frac{7500}{5208} \approx 1.44^{\circ} \mathrm{C} $$
Therefore, the rise in temperature is $1.44^{\circ} \mathrm{C}$.
Hence, the correct option is (B).

Answer

1.75

Answer

1.44

Answer

1.95

Answer

1.15

Centre of Mass and Rotational Motion

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