Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively. Sphere A, of mass 2 kg, collides directly with sphere B, of mass 5 kg. The coefficient of restitution for the collision is rac{2}{5}. Find (i) the speeds of A and B immediately after the collision (ii) the loss of kinetic energy due to the collision (iii) the magnitude of the impulse imparted to B due to the collision. A ball is fired vertically upward in a room with a smooth horizontal floor and a smooth horizontal ceiling. The height of the room is 3 metres. The ball is fired upwards at a speed of 12 m s⁻¹ from a height of 1 metre above the floor. The coefficient of restitution for all collisions between the ball and the ceiling and between the ball and the floor is rac{3}{5}. (i) Find the speed of the ball immediately after striking the ceiling. (ii) Investigate whether the ball strikes the ceiling again after rebounding from the floor.

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Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively - Leaving Cert Applied Maths - Question 5 - 2019

Question 5

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Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively. Sphere A, of mass 2 kg, col... show full transcript

Worked Solution & Example Answer:Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively - Leaving Cert Applied Maths - Question 5 - 2019

Step 1

(i) the speeds of A and B immediately after the collision

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Answer

Using the principle of conservation of momentum:

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$

Substituting the values:

$2(3) + 5(-1) = 2 v_A + 5 v_B$

This simplifies to:

$6 - 5 = 2 v_A + 5 v_B$ $1 = 2 v_A + 5 v_B \quad (1)$

For the coefficient of restitution (e) given by:

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

Substituting the initial and final velocities:

$\frac{v_B - v_A}{3 - (-1)} = \frac{2}{5}$

This becomes:

$5(v_B - v_A) = 2(4)$ $5(v_B - v_A) = 8$ $v_B - v_A = 1.6 \quad (2)$

Now, we can solve equations (1) and (2) simultaneously. From equation (2):

$v_B = v_A + 1.6$

Substituting this into equation (1):

$1 = 2 v_A + 5(v_A + 1.6)$ $1 = 2 v_A + 5 v_A + 8$ $1 = 7 v_A + 8$ $7 v_A = -7$ $v_A = -1 \text{ m s}^{-1}$

Plugging back to find $v_B$ :

$v_B = -1 + 1.6 = 0.6 \text{ m s}^{-1}$

Thus, the speeds are:

Speed of A: -1 m/s (indicating a change in direction)
Speed of B: 0.6 m/s

Step 2

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

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Answer

The kinetic energy before the collision is given by: $KE_{initial} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2$

Calculating:

$KE_{initial} = \frac{1}{2}(2)(3)^2 + \frac{1}{2}(5)(1)^2 = \frac{1}{2}(2)(9) + \frac{5}{2} = 9 + 2.5 = 11.5 \text{ J}$

The kinetic energy after the collision is:

$KE_{final} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2$

Calculating:

$KE_{final} = \frac{1}{2}(2)(-1)^2 + \frac{1}{2}(5)(0.6)^2 = \frac{1}{2}(2)(1) + \frac{1}{2}(5)(0.36) = 1 + 0.9 = 1.9 \text{ J}$

Now, the loss of kinetic energy is:

$\text{Loss} = KE_{initial} - KE_{final} = 11.5 - 1.9 = 9.6 \text{ J}$

Step 3

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

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Answer

Impulse can be calculated using the change in momentum:

$I = m_B (v_B - u_B)$

Substituting in the values:

$I = 5(0.6 - (-1)) = 5(0.6 + 1) = 5(1.6) = 8 \text{ kg m/s}$

Thus, the magnitude of the impulse imparted to B is 8 kg m/s.

Step 4

(i) Find the speed of the ball immediately after striking the ceiling.

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Answer

Using the kinematic equation:

$v^2 = u^2 + 2as$

Here, the initial velocity is 12 m/s, a = -g (where g is the acceleration due to gravity, approximately 10 m/s²) and the displacement is 2 m:

$v^2 = 12^2 + 2(-10)(2)$

Calculating:

$v^2 = 144 - 40 = 104$

So,

$v = \sqrt{104} = 10.2 \text{ m/s}$

Thus, the speed of the ball immediately after striking the ceiling is approximately 10.2 m/s.

Step 5

(ii) Investigate whether the ball strikes the ceiling again after rebounding from the floor.

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Answer

Let's find the speed just before the ball strikes the floor: Using the kinematic equation again for its descent:

$v^2 = u^2 + 2as$

Where the initial speed at the peak is 0 m/s and the total height is 2 m:

ightarrow v = \sqrt{40} = 6.32 \text{ m/s}$$ Now calculating the rebound speed using the coefficient of restitution: Let the speed just after striking the floor be $v' = e v$ where $e = \frac{3}{5}$: $$v' = \frac{3}{5} (6.32) = 3.8 \text{ m/s}$$ The maximum height reached after rebounding can be found: Using: $$v'^2 = u'^2 + 2as$$ Setting final velocity to 0 at the peak: $$0 = (3.8)^2 + 2(-10)s$$ Solving for s: $$s = \frac{(3.8)^2}{20} = \frac{14.44}{20} = 0.722 \text{ m}$$ The height after rebounding is: Height = 1 m (initial height) + 0.722 m = 1.722 m. Since ceiling is at 3 m, it does NOT strike the ceiling again.

Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively - Leaving Cert Applied Maths - Question 5 - 2019

Worked Solution & Example Answer:Two smooth spheres A and B are sliding towards each other on a smooth horizontal table, with speeds of 3 m s⁻¹ and 1 m s⁻¹, respectively - Leaving Cert Applied Maths - Question 5 - 2019

(i) the speeds of A and B immediately after the collision

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

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

(i) Find the speed of the ball immediately after striking the ceiling.

(ii) Investigate whether the ball strikes the ceiling again after rebounding from the floor.

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