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A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 5 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2008
Question 5
A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 5 kg, on a smooth horizontal table. A and B are moving in opposite directio... show full transcript
Worked Solution & Example Answer:A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 5 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2008
Step 1
(i) the speed of A and the speed of B after the collision
Answer
To find the speeds after the collision, we apply the principle of conservation of momentum (PCM) and the principle of conservation of energy. The equation for PCM is:
[ 6(4) + 5(2) = 6v_1 + 5v_2 ]
Where:\n- ( v_1 ) is the final velocity of A\n- ( v_2 ) is the final velocity of B \n
Solving this gives:
[ 24 + 10 = 6v_1 + 5v_2 ]
[ 34 = 6v_1 + 5v_2 ] ( \text{(1)} )
Next, we apply the coefficient of restitution (COR), given as ( \frac{1}{10} ), which relates the velocities before and after the collision: [ e = \frac{u_2 - u_1}{v_1 - v_2} ] [ \frac{1}{10} = \frac{(2 - (-4))}{(v_1 - v_2)} ] ( \text{(2)} )
From (2), we can rearrange to find: [ v_1 - v_2 = 10] ( \text{(3)} )
Now, combine equations (1) and (3):
From (3): ( v_2 = v_1 - 10 )
Plugging this back into (1):
[ 34 = 6v_1 + 5(v_1 - 10) ]
[ 34 = 6v_1 + 5v_1 - 50 ]
[ 34 + 50 = 11v_1 ]
[ 84 = 11v_1 ]
[ v_1 = \frac{84}{11} = 7.64 , m/s ]
[ v_2 = v_1 - 10 = 7.64 - 10 = -2.36 , m/s ]
Thus, the speeds after collision are:
- Speed of A: ( 7.64 , m/s )\n- Speed of B: ( -2.36 , m/s ) (direction change indicates opposite movement).
Step 2
(ii) the loss in kinetic energy due to the collision
Answer
The kinetic energy before and after the collision can be calculated using the formula: [ KE = \frac{1}{2} mv^2 ]
Calculate KE before the collision: [ KE_{before} = \frac{1}{2} (6)(4^2) + \frac{1}{2} (5)(2^2) ] [ = \frac{1}{2} (6)(16) + \frac{1}{2} (5)(4) ] [ = 48 + 10 = 58 , J ]
Calculate KE after the collision using the speeds found: [ KE_{after} = \frac{1}{2} (6)(7.64^2) + \frac{1}{2} (5)(-2.36^2) ] [ = \frac{1}{2} (6)(58.0336) + \frac{1}{2} (5)(5.5696) ] [ = 174.1008 + 13.924 = 188.0248 , J ]
Thus, the loss in kinetic energy is calculated as follows: [ KE_{lost} = KE_{before} - KE_{after} = 58 - 188.0248 = -130.0248 , J ]
So, the loss of kinetic energy due to the collision is ( 130.0248 , J ).
Step 3
(iii) the magnitude of the impulse imparted to A due to the collision
Answer
Impulse can be calculated using the formula: [ Impulse = \Delta p = m(v_final - v_initial) ]
Where:
- Mass of A, ( m = 6 , kg )\n- Final velocity of A, ( v_{final} = 7.64 , m/s )\n- Initial velocity of A, ( v_{initial} = 4 , m/s )
Thus, the impulse is calculated as follows: [ Impulse = 6(7.64 - 4) = 6(3.64) = 21.84 , Ns ]
Therefore, the magnitude of the impulse imparted to A due to the collision is ( 21.84 , Ns ).