Two small smooth spheres A and B, of equal radii and of masses km kg and m kg respectively, where k > 1, are free to move on a smooth horizontal plane. A is moving towards B with speed 6 m s−1 and B is moving towards A with speed 2 m s−1. After the collision A and B coalesce and move with speed 4 m s−1.
(a) Question
Find k.
▶️Answer/Explanation
(a) Finding \(k\):
Using the conservation of linear momentum:
\( km \times 6 – m \times 2 = (km + m) \times 4 \)
\( k = 3 \)
(b) Question
Find, in terms of m, the loss of kinetic energy due to the collision.
▶️Answer/Explanation
(b) Finding the Loss of Kinetic Energy:
The initial kinetic energy (\(KE_{\text{initial}}\)) is the sum of the kinetic energies of A and B:
\( KE_{\text{initial}} = \frac{1}{2} \times k m \times 6^2 + \frac{1}{2} \times m \times (-2)^2 \)
The final kinetic energy (\(KE_{\text{final}}\)) is the kinetic energy of the coalesced mass:
\( KE_{\text{final}} = \frac{1}{2} \times (k m + m) \times 4^2 \)
The loss of kinetic energy (\(\Delta KE\)) is given by:
\(\Delta KE = KE_{\text{initial}} – KE_{\text{final}} \)
Substitute the values for \(k\) and \(m\):
\( \Delta KE = 24 \, \text{m} \, \text{J} \)
Question
A particle B of mass 5 kg is at rest on a smooth horizontal table. A particle A of mass 2.5 kg moves on the table with a speed of 6 m s−1 and collides directly with B. In the collision the two particles coalesce.
(a) Find the speed of the combined particle after the collision.
(b) Find the loss of kinetic energy of the system due to the collision.
▶️Answer/Explanation
(a) Finding the Speed of the Combined Particle after the Collision:
Using the conservation of linear momentum:
\( m_A \cdot v_A + m_B \cdot v_B = (m_A + m_B) \cdot v_f \)
Substituting the given values:
\( 6 \times 2.5 + 5 \times 0 = (2.5 + 5) \times v_f \)
Solving for \(v_f\):
\( v_f = \frac{15}{7} \, \text{m/s} \)
(b) Finding the Loss of Kinetic Energy:
Using the kinetic energy formula \(KE = \frac{1}{2} m v^2\) either before or after the collision:
Before the collision:
\( KE_{\text{initial}} = 0.5 \times 2.5 \times 6^2 \)
After the collision:
\( KE_{\text{final}} = 0.5 \times (2.5 + 5) \times \left(\frac{15}{7}\right)^2 \)
The loss of kinetic energy (\(\Delta KE\)) is given by:
\( \Delta KE = KE_{\text{initial}} – KE_{\text{final}} \)
Now, substitute the values and calculate \(\Delta KE\):
\( \Delta KE = 30 \, \text{J} \)