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A smooth sphere A, of mass 3 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2011
Question 5
A smooth sphere A, of mass 3 kg, collides directly with another smooth sphere B, of mass 1 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 3 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2011
Step 1
Find (i) the speed of A and the speed of B after the collision
Answer
To find the speeds of A and B after the collision, we use the principles of conservation of momentum and the coefficient of restitution.
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Conservation of Momentum: The total momentum before the collision equals the total momentum after the collision.
[ 3(5) + 1(-2) = 3v_A + 1v_B ]
[ 15 - 2 = 3v_A + v_B ]
[ 13 = 3v_A + v_B , \quad (1) ]
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Coefficient of Restitution: This relates the velocities before and after the collision.
[ \frac{v_B - v_A}{5 + 2} = \frac{1}{7} ]
[ v_B - v_A = \frac{1}{7} \times 7 \quad \Rightarrow \quad v_B - v_A = 1 \quad (2) ]
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Solving the equations: Now we have two equations.
From (2): [ v_B = v_A + 1 ]
Substitute ( v_B ) in (1):
[ 13 = 3v_A + (v_A + 1) ]
[ 13 = 4v_A + 1 ]
[ 12 = 4v_A ]
[ v_A = 3 \text{ ms}^{-1} ]
Now substituting ( v_A ) back into (2):
[ v_B = 3 + 1 = 4 \text{ ms}^{-1} ]
So, the speeds after the collision are:
( v_A = 3 , \text{ms}^{-1} ) and ( v_B = 4 , \text{ms}^{-1} ).
Step 2
Find (ii) the loss in kinetic energy due to the collision
Answer
Now we calculate the kinetic energies before and after the collision.
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Initial Kinetic Energy (KE):
[ KE_i = \frac{1}{2} \times 3 \times (5)^2 + \frac{1}{2} \times 1 \times (2)^2 ]
[ KE_i = \frac{1}{2} \times 3 \times 25 + \frac{1}{2} \times 1 \times 4 ]
[ KE_i = 39.5 , \text{J} ]
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Final Kinetic Energy (KE):
[ KE_f = \frac{1}{2} \times 3 \times (3)^2 + \frac{1}{2} \times 1 \times (4)^2 ]
[ KE_f = \frac{1}{2} \times 3 \times 9 + \frac{1}{2} \times 1 \times 16 ]
[ KE_f = 21.5 , \text{J} ]
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Loss in Kinetic Energy:
[ Loss = KE_i - KE_f ]
[ Loss = 39.5 - 21.5 = 18 , \text{J} ]
Thus, the loss in kinetic energy due to the collision is 18 J.
Step 3
Find (iii) the magnitude of the impulse imparted to B due to the collision
Answer
Finally, we calculate the impulse imparted to B by using the formula:
[ I = m(v_B - u_B) ]
Where:
- ( m ) is mass of B = 1 kg
- ( v_B ) is the final velocity of B = 4 ms⁻¹
- ( u_B ) is the initial velocity of B = -2 ms⁻¹ (opposite direction)
[ I = 1(4 - (-2)) = 1(4 + 2) = 1 imes 6 = 6 , \text{Ns} ]
Therefore, the magnitude of the impulse imparted to B due to the collision is 6 Ns.