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):

• A. $1.24^\circ$C
• B. $2.44^\circ$C
• C. $2.24^\circ$C
• D. $1.44^\circ$C

Hint:

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.

Answer 1

The Correct Option is D

To find the rise in temperature of the balls, we need to calculate the energy lost during the perfectly inelastic collision and how that energy translates into heat to raise the temperature of each ball.

First, let's analyze the given collision:

The collision is perfectly inelastic, meaning the two balls stick together after the collision.

The relevant formula for a perfectly inelastic collision is:

\(v = \frac{{m_1 v_1 + m_2 v_2}}{{m_1 + m_2}}\)

• \(m_1 = 15 \, \text{kg}\)
• \(v_1 = 10 \, \text{m/s}\)
• \(m_2 = 25 \, \text{kg}\)
• \(v_2 = -30 \, \text{m/s}\)

Calculate the final velocity \(v\):

\(v = \frac{(15 \times 10) + (25 \times (-30))}{15 + 25} = \frac{150 - 750}{40} = -15 \, \text{m/s}\)

The velocity is negative, indicating a reversal in the direction.

Next, calculate the initial kinetic energy (KE) of the system:

\(\text{KE}_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\)

Substitute the values:

\(\text{KE}_{\text{initial}} = \frac{1}{2} \times 15 \times 10^2 + \frac{1}{2} \times 25 \times (-30)^2\)

\(\text{KE}_{\text{initial}} = 750 + 11250 = 12000 \, \text{J}\)

Calculate the final kinetic energy (\(\text{KE}_{\text{final}}\)):

\(\text{KE}_{\text{final}} = \frac{1}{2} \times 40 \times (-15)^2 = 4500 \, \text{J}\)

Energy lost in the collision:

\(\text{Energy lost} = \text{KE}_{\text{initial}} - \text{KE}_{\text{final}} = 12000 - 4500 = 7500 \, \text{J}\)

The energy lost is converted to heat, increasing the temperature. Using the heat-energy relation:

 

Convert joules to calories (1 cal = 4.184 J):

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