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

To determine the rise in temperature during a perfectly inelastic collision where two bodies stick together, we need to consider the conservation of momentum and the conversion of kinetic energy into thermal energy.

Step 1: Calculate Initial Momentum

Consider two spheres with masses \( m_1 = 15 \, \text{kg} \) and \( m_2 = 25 \, \text{kg} \). Their initial velocities are \( v_1 = 10 \, \text{m/s} \) and \( v_2 = -30 \, \text{m/s} \) (since they are moving in opposite directions).

The total initial momentum (\( p_i \)) is given by:

\(p_i = m_1 \cdot v_1 + m_2 \cdot v_2 = 15 \times 10 + 25 \times (-30) = 150 - 750 = -600 \, \text{kg m/s}\)

Step 2: Calculate Final Momentum and Velocity

In a perfectly inelastic collision, the spheres stick together. Let \( v_f \) be their final velocity. The final momentum (\( p_f \)) is:

\(p_f = (m_1 + m_2) v_f = 40 v_f \, \text{kg m/s}\)

From the conservation of momentum, \( p_i = p_f \):

\(-600 = 40 v_f \Rightarrow v_f = -15 \, \text{m/s}\)

Step 3: Calculate Initial Kinetic Energy

The initial kinetic energy (\( KE_i \)) is:

\(KE_i = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\) \(KE_i = \frac{1}{2} \times 15 \times (10)^2 + \frac{1}{2} \times 25 \times (30)^2\) \(KE_i = 750 + 11250 = 12000 \, \text{J}\)

Step 4: Calculate Final Kinetic Energy

The final kinetic energy (\( KE_f \)) is:

\(KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 40 \times (15)^2 = 4500 \, \text{J}\)

Step 5: Calculate Heat Produced

The heat produced as a result of energy loss is:

\(\Delta KE = KE_i - KE_f = 12000 - 4500 = 7500 \, \text{J}\)

Step 6: Calculate Rise in Temperature

Convert joules into calories, using \( 1 \, \text{cal} = 4.2 \, \text{J} \):

\(\Delta Q = \frac{7500}{4.2} = 1785.71 \, \text{cal}\)

The rise in temperature (\( \Delta T \)) using specific heat \( c = 31 \, \text{cal/kg°C} \) is given by:

\(\Delta T = \frac{\Delta Q}{(m_1 + m_2) \cdot c} = \frac{1785.71}{40 \times 31} \approx 1.44^\circ \text{C}\)

Thus, the rise in temperature is 1.44°C. Therefore, the correct answer is 1.44.