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 m/s and 30 m/s, respectively, strike each other and stick together. The rise in temperature (in °C), if all the heat produced during the collision is retained by these spheres, is (specific heat 31 cal/kg.°C and 1 cal = 4.2 J) :

• A. 1.95
• B. 1.15
• C. 1.44
• D. 1.75

Hint:

Remember to convert specific heat from calories to Joules before equating it to the kinetic energy loss.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a perfectly inelastic collision, kinetic energy is lost. This lost energy is converted into heat. We use conservation of momentum to find the final velocity, then calculate the energy loss and relate it to temperature rise using calorimetry.
Step 2: Key Formula or Approach:
1. Momentum: \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\).
2. Heat: \(Q = \Delta K.E. = (m_1+m_2)s\Delta T\).
Step 3: Detailed Explanation:
Let \(m_1 = 15\) kg, \(u_1 = 10\) m/s and \(m_2 = 25\) kg, \(u_2 = -30\) m/s.
\[ 15(10) + 25(-30) = (15+25)v \implies 150 - 750 = 40v \implies v = -15 \text{ m/s} \] Initial K.E. \( = \frac{1}{2}(15)(10)^2 + \frac{1}{2}(25)(30)^2 = 750 + 11250 = 12000\) J.
Final K.E. \( = \frac{1}{2}(40)(15)^2 = 20 \times 225 = 4500\) J.
Heat produced \(Q = 12000 - 4500 = 7500\) J.
Specific heat \(s = 31 \times 4.2 = 130.2\) J/kg°C.
\[ 7500 = (40)(130.2)\Delta T \] \[ \Delta T = \frac{7500}{5208} \approx 1.44^\circ \text{C} \]
Step 4: Final Answer:
The rise in temperature is 1.44 °C.