A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal table. A and B are moving in the same direction with speeds of 4 m s⁻¹ and 1 m s⁻¹ respectively. The coefficient of restitution for the collision is $ \frac{1}{6} $ Find (i) the speed of A and the speed of B after the collision (ii) the loss in kinetic energy 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 5 kg, collides directly with another smooth sphere B, of mass 2 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2012

Question 5

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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. A and B are moving in the same directio... 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 - 2012

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 spheres A and B after the collision, we can use the principle of conservation of momentum along with the coefficient of restitution.

Step 1: Conservation of Momentum

The total momentum before the collision can be expressed as:

$5(4) + 2(1) = 5v_A + 2v_B$

Where:

5 and 2 are the masses of spheres A and B, respectively.
4 and 1 are their initial speeds.
(v_A) and (v_B) are the final speeds after collision.

Step 2: Coefficient of Restitution

The coefficient of restitution (e) is given by:

$e = \frac{v_B - v_A}{u_A - u_B}$

Here, (u_A = 4) m/s and (u_B = 1) m/s (initial speeds of A and B). Substituting the given value of e:

$\frac{1}{6} = \frac{v_B - v_A}{4 - 1}$

Which simplifies to:

$v_B - v_A = \frac{1}{6} \times 3 = \frac{1}{2}$

From these two equations, we get:

$20 = 5v_A + 2v_B$
$v_B = v_A + \frac{1}{2}$

Now, substituting equation 2 into equation 1 yields:

$20 = 5v_A + 2(v_A + \frac{1}{2})$

This results in:

$20 = 7v_A + 1$ $19 = 7v_A \Rightarrow v_A = \frac{19}{7} \approx 2.71 \text{ m/s}$

Substituting (v_A) back to find (v_B):

$v_B = v_A + \frac{1}{2} = \frac{19}{7} + \frac{1}{2} \approx 3.21 \text{ m/s}$

Thus, the final speeds after collision are: $v_A \approx 2.71 \text{ m/s}, \quad v_B \approx 3.21 \text{ m/s}$

Step 2

(ii) the loss in kinetic energy due to the collision

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Answer

The kinetic energy (KE) before and after the collision can be computed as follows:

Step 1: Initial Kinetic Energy

The initial kinetic energy of the system is given by:

$KE_{initial} = \frac{1}{2} (5)(4^2) + \frac{1}{2} (2)(1^2)$ $KE_{initial} = \frac{1}{2} (5)(16) + \frac{1}{2} (2)(1) = 40 + 1 = 41 \text{ J}$

Step 2: Final Kinetic Energy

Now for the final kinetic energy after the collision:

$KE_{final} = \frac{1}{2} (5)(v_A^2) + \frac{1}{2} (2)(v_B^2)$

Substituting (v_A \approx 2.71) and (v_B \approx 3.21):

$KE_{final} = \frac{1}{2} (5)(2.71^2) + \frac{1}{2} (2)(3.21^2) \approx \frac{1}{2} (5)(7.34) + \frac{1}{2} (2)(10.30) \approx 18.35 + 10.30 = 28.65 \text{ J}$

Step 3: Loss in Kinetic Energy

The loss in kinetic energy is:

$KE_{loss} = KE_{initial} - KE_{final} = 41 - 28.65 \approx 12.35 \text{ J}$

Step 3

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

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Answer

The impulse imparted to an object can be calculated using the formula:

$Impulse = \Delta p = m(v_{final} - v_{initial})$

Where m is the mass of object A and (v_{final} - v_{initial}) is the change in velocity.

Step 1: Calculate Change in Momentum

For sphere A:

Mass (m_A = 5 \text{ kg} )
Initial velocity (u_A = 4 \text{ m/s} )
Final velocity (v_A = 2.71 \text{ m/s} )

Thus, the change in velocity is:

$\Delta v_A = v_A - u_A = 2.71 - 4 = -1.29 \text{ m/s}$

Step 2: Calculate Impulse

Now substituting into the impulse formula:

$Impulse = 5 \times (-1.29) = -6.45 \text{ Ns}$

Taking the magnitude:

$|Impulse| = 5 \text{ Ns}$

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 - 2012

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 - 2012

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

(ii) the loss in kinetic energy due to the collision

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

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