A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table. A and B are moving in the same direction with speeds of 5 ms⁻¹ and 4 ms⁻¹ respectively. The coefficient of restitution for the collision is \rac{2}{3}. Find (i) the speed of A and the speed of B after the collision (ii) the change in the kinetic energy of A due to the collision (iii) the magnitude of the impulse imparted to A due to the collision.

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A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2010

Question 5

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A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table. A and B are moving in the same directio... show full transcript

Worked Solution & Example Answer:A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2010

Step 1

(i) the speed of A and the speed of B after the collision

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Answer

To find the speeds of A and B after the collision, we apply the principles of conservation of momentum and the coefficient of restitution.

Conservation of Momentum (PCM):

The equation can be expressed as:

$2(s) + 3(4) = 2(v_A) + 3(v_B)$

Substituting the initial velocities:

$2(5) + 3(4) = 2(v_A) + 3(v_B)$

This simplifies to:

22 = 2v_A + 3v_B \ \ (1)$$ Next, we apply the **Newton's Law of Restitution (NEL)**: $$v_A - v_B = -e (u_A - u_B)$$ Where: - $u_A = 5 ms^{-1}$, $u_B = 4 ms^{-1}$ - e (coefficient of restitution) = \ rac{2}{3} Substituting these values gives: $$v_A - v_B = -\frac{2}{3}(5 - 4)\ = -\frac{2}{3}\ (2)$$ Now from (1) and (2), we can solve the two equations simultaneously. Substituting (2) into (1): From (2): $v_A = v_B - \frac{2}{3}$ Substituting: $$2((v_B - \frac{2}{3})) + 3v_B = 22\ 2v_B - \frac{4}{3} + 3v_B = 22\ 5v_B - \frac{4}{3} = 22\ 5v_B = 22 + \frac{4}{3}\ 5v_B = \frac{66 + 4}{3}\ 5v_B = \frac{70}{3}\ v_B = \frac{14}{3}\ v_B = 4.67 ms^{-1} \ v_A = 4 ms^{-1}$$

Step 2

(ii) the change in the kinetic energy of A due to the collision

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Answer

To find the change in the kinetic energy of sphere A, we calculate the kinetic energy before and after the collision.

Kinetic Energy Before Collision:

The kinetic energy for mass A before the collision is given by:

$KE_{before} = \frac{1}{2} m v^2 = \frac{1}{2} (2)(5^2) = 25 J$

Kinetic Energy After Collision:

Using the new speed of A:

$KE_{after} = \frac{1}{2} m v_A^2 = \frac{1}{2} (2)(4^2) = 16 J$

Change in Kinetic Energy:

$\Delta KE = KE_{after} - KE_{before} = 16 - 25 = -9 J$

Step 3

(iii) the magnitude of the impulse imparted to A due to the collision

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Answer

The impulse imparted to A is calculated using the change in momentum:

Impulse:

$Impulse = m(v_A - u_A)$

Where:

$u_A$ = initial speed = 5 ms⁻¹
$v_A$ = speed after collision = 4 ms⁻¹

Substituting the values gives:

$Impulse = 2(4 - 5) = 2(-1) = -2 N s$

Taking the magnitude, we get:

$|Impulse| = 2 N s$

A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2010

Worked Solution & Example Answer:A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2010

(i) the speed of A and the speed of B after the collision

(ii) the change in the kinetic energy of A due to the collision

(iii) the magnitude of the impulse imparted to A due to the collision

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