Question:
Two balls made of the same material collide perfectly inelastically as shown. The energy lost in the collision is completely utilized in raising the temperature of each ball. Find the rise in temperature of the balls. (Specific heat $=31$ cal/kg-$^\circ$C): 
Two balls made of the same material collide perfectly inelastically as shown. The energy lost in the collision is completely utilized in raising the temperature of each ball. Find the rise in temperature of the balls. (Specific heat $=31$ cal/kg-$^\circ$C):
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In perfectly inelastic collisions, always find energy loss using kinetic energies before and after collision.
If objects are of the same material, divide total heat gained by total heat capacity to get temperature rise.
Updated On: Jan 31, 2026
- $1.24^\circ$C
- $2.44^\circ$C
- $2.24^\circ$C
- $1.44^\circ$C
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The Correct Option is D
Solution and Explanation
Concept:
In a perfectly inelastic collision:
Linear momentum is conserved
Loss of kinetic energy appears as internal energy (heat)
If bodies are of the same material, they attain the same rise in temperature
Step 1: Given data Mass of ball $A$: \[ m_1 = 15\ \text{kg}, \quad v_1 = +10\ \text{m/s} \] Mass of ball $B$: \[ m_2 = 25\ \text{kg}, \quad v_2 = -30\ \text{m/s} \] Specific heat: \[ c = 31\ \text{cal/kg-}^\circ\text{C} \]
Step 2: Find common velocity after collision Using conservation of momentum: \[ m_1v_1 + m_2v_2 = (m_1+m_2)v \] \[ 15(10) + 25(-30) = 40v \] \[ 150 - 750 = 40v \Rightarrow v = -15\ \text{m/s} \]
Step 3: Initial kinetic energy \[ K_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \] \[ K_i = \frac{1}{2}(15)(10^2) + \frac{1}{2}(25)(30^2) = 750 + 11250 = 12000\ \text{J} \] Step 4: Final kinetic energy \[ K_f = \frac{1}{2}(40)(15^2) = 4500\ \text{J} \]
Step 5: Energy lost in collision \[ \Delta K = K_i - K_f = 12000 - 4500 = 7500\ \text{J} \] Convert into calories: \[ Q = \frac{7500}{4.2} \approx 1786\ \text{cal} \]
Step 6: Rise in temperature Total heat capacity of both balls: \[ (m_1+m_2)c = 40 \times 31 = 1240\ \text{cal/}^\circ\text{C} \] \[ \Delta T = \frac{Q}{(m_1+m_2)c} = \frac{1786}{1240} \approx 1.44^\circ\text{C} \] Conclusion: The rise in temperature of each ball is: \[ \boxed{1.44^\circ\text{C}} \]
Linear momentum is conserved
Loss of kinetic energy appears as internal energy (heat)
If bodies are of the same material, they attain the same rise in temperature
Step 1: Given data Mass of ball $A$: \[ m_1 = 15\ \text{kg}, \quad v_1 = +10\ \text{m/s} \] Mass of ball $B$: \[ m_2 = 25\ \text{kg}, \quad v_2 = -30\ \text{m/s} \] Specific heat: \[ c = 31\ \text{cal/kg-}^\circ\text{C} \]
Step 2: Find common velocity after collision Using conservation of momentum: \[ m_1v_1 + m_2v_2 = (m_1+m_2)v \] \[ 15(10) + 25(-30) = 40v \] \[ 150 - 750 = 40v \Rightarrow v = -15\ \text{m/s} \]
Step 3: Initial kinetic energy \[ K_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \] \[ K_i = \frac{1}{2}(15)(10^2) + \frac{1}{2}(25)(30^2) = 750 + 11250 = 12000\ \text{J} \] Step 4: Final kinetic energy \[ K_f = \frac{1}{2}(40)(15^2) = 4500\ \text{J} \]
Step 5: Energy lost in collision \[ \Delta K = K_i - K_f = 12000 - 4500 = 7500\ \text{J} \] Convert into calories: \[ Q = \frac{7500}{4.2} \approx 1786\ \text{cal} \]
Step 6: Rise in temperature Total heat capacity of both balls: \[ (m_1+m_2)c = 40 \times 31 = 1240\ \text{cal/}^\circ\text{C} \] \[ \Delta T = \frac{Q}{(m_1+m_2)c} = \frac{1786}{1240} \approx 1.44^\circ\text{C} \] Conclusion: The rise in temperature of each ball is: \[ \boxed{1.44^\circ\text{C}} \]
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