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A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2009
Question 5
A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal table. Before impact A and B are moving in opp... show full transcript
Worked Solution & Example Answer:A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2009
Step 1
(i) the speed of A and the speed of B after the collision
Answer
To determine the speeds of A and B after the collision, we can apply the principles of conservation of momentum (PCM) and the coefficient of restitution (COR).
1. Conservation of Momentum: Using the equation for momentum before and after the collision, we have:
[ m_A u_A + m_B u_B = m_A v_A + m_B v_B ]
Where:
- ( m_A = 5 , kg )
- ( m_B = 2 , kg )
- ( u_A = 3 , m/s ) (initial speed of A)
- ( u_B = -5 \ ,m/s ) (initial speed of B)
- ( v_A ) and ( v_B ) are the final speeds of A and B, respectively.
Substituting the known values: [ 5(3) + 2(-5) = 5v_A + 2v_B ] [ 15 - 10 = 5v_A + 2v_B ] [ 5 = 5v_A + 2v_B \quad (1) ]
2. Coefficient of Restitution: The formula relating relative velocities before and after the collision is given by: [ e = \frac{v_B - v_A}{u_A - u_B} ]
Where:
- ( e = \frac{3}{4} )
- The relative speeds before impact are ( 3 - (-5) = 3 + 5 = 8 ).
Substituting: [ \frac{3}{4} = \frac{v_B - v_A}{8} ]
From which we have: [ 3 = 4(v_B - v_A) ] [ v_B - v_A = \frac{3}{4} \quad (2) ]
3. Solving the Equations: We now have two equations: (1) and (2). From (2): [ v_B = v_A + \frac{3}{4} ]
Substituting into (1): [ 5 = 5v_A + 2(v_A + \frac{3}{4}) ] [ 5 = 5v_A + 2v_A + \frac{3}{2} ] [ 5 = 7v_A + 1.5 ] [ 7v_A = 5 - 1.5 ] [ 7v_A = 3.5 ] [ v_A = \frac{3.5}{7} = \frac{1}{2} \text{ m/s} ]
Now we can find ( v_B ): [ v_B = \frac{1}{2} + \frac{3}{4} = \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4} \text{ m/s} ]
Final Speeds:
- The speed of A after the collision: ( 0.5 , m/s )
- The speed of B after the collision: ( 1.25 , m/s )
Step 2
(ii) the loss in kinetic energy due to the collision
Answer
To calculate the loss in kinetic energy, we first determine the kinetic energy before and after the collision.
1. Kinetic Energy Before Collision: [ KE_{before} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 ] Substituting the values: [ KE_{before} = \frac{1}{2} \times 5 \times (3)^2 + \frac{1}{2} \times 2 \times (-5)^2 ] [ KE_{before} = \frac{1}{2} \times 5 \times 9 + \frac{1}{2} \times 2 \times 25 ] [ KE_{before} = 22.5 + 25 = 47.5 , J ]
2. Kinetic Energy After Collision: [ KE_{after} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 ] Substituting the final speeds: [ KE_{after} = \frac{1}{2} \times 5 \times \left( \frac{1}{2} \right)^2 + \frac{1}{2} \times 2 \times \left( \frac{5}{4} \right)^2 ] [ KE_{after} = \frac{1}{2} \times 5 \times \frac{1}{4} + \frac{1}{2} \times 2 \times \frac{25}{16} ] [ KE_{after} = \frac{5}{8} + \frac{25}{16} ] [ KE_{after} = \frac{10}{16} + \frac{25}{16} = \frac{35}{16} = 27.5 , J ]
3. Loss in Kinetic Energy: [ KE \text{ lost} = KE_{before} - KE_{after} ] [ KE \text{ lost} = 47.5 - 27.5 = 20 , J ]
Step 3
(iii) the magnitude of the impulse imparted to B due to the collision
Answer
Impulse is defined as the change in momentum. The formula for impulse imparted to object B is:
[ Impulse = m_B (v_B - u_B) ]
Where:
- ( m_B = 2 , kg )
- ( v_B = \frac{5}{4} , m/s )
- ( u_B = -5 , m/s )
Substituting the known values: [ Impulse = 2 \left( \frac{5}{4} - (-5) \right) ] [ Impulse = 2 \left( \frac{5}{4} + 5 \right) ] [ Impulse = 2 \left( \frac{5}{4} + \frac{20}{4} \right) = 2 \left( \frac{25}{4} \right) ] [ Impulse = \frac{50}{4} = 12.5 , Ns ]
Thus, the magnitude of the impulse imparted to B is ( 20 , Ns ).