a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table. Spheres A and B are moving in opposite directions with speeds of 2 m s⁻¹ and 3 m s⁻¹ respectively. The coefficient of restitution for the collision is $rac{3}{5}$. 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. b) A ball is fired vertically down with a speed of 2 m s⁻¹ from a height of 3 metres onto a smooth horizontal floor. The ball hits the floor and rebounds to a height of 1.8 metres. The coefficient of restitution between the ball and the floor is e. Find (i) the speed of the ball when it hits the floor (ii) the value of e.

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a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2016

Question 5

a)-A-smooth-sphere-A,-of-mass-6-kg,-collides-directly-with-another-smooth-sphere-B,-of-mass-4-kg,-on-a-smooth-horizontal-table-Leaving Cert Applied Maths-Question 5-2016.png

a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table. Spheres A and B are moving in opposi... show full transcript

Worked Solution & Example Answer:a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2016

Step 1

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

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Answer

To find the final speeds of spheres A and B after the collision, we can use the equations of momentum and the coefficient of restitution.

First, let's denote:

Initial speed of A, $u_A = 2 ext{ m/s}$
Initial speed of B, $u_B = -3 ext{ m/s}$ (B is moving in the opposite direction)
Mass of A, $m_A = 6 ext{ kg}$
Mass of B, $m_B = 4 ext{ kg}$

Using conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$ Where $v_A$ and $v_B$ are the final speeds of A and B respectively.

Substituting the known values: $6(2) + 4(-3) = 6v_A + 4v_B$ $12 - 12 = 6v_A + 4v_B$ $0 = 6v_A + 4v_B$ $6v_A = -4v_B$ $v_A = - rac{2}{3}v_B$

From the coefficient of restitution: $e = rac{v_B - v_A}{u_A - u_B}$ Substituting $e = rac{3}{5}$ : $rac{3}{5} = rac{v_B - (- rac{2}{3}v_B)}{2 - 3}$ Solving gives: $rac{3}{5} = rac{v_B + rac{2}{3}v_B}{-1} = - rac{5}{3}v_B$ Thus, $v_B = - rac{9}{5} ext{ m/s}$ Then, substituting back to find $v_A$ : $v_A = - rac{2}{3}(- rac{9}{5}) = rac{6}{5} ext{ m/s}$

Therefore, the final speeds are:

Speed of A after collision: $v_A = rac{6}{5} ext{ m/s}$
Speed of B after collision: $v_B = - rac{9}{5} ext{ m/s}$

Step 2

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

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Answer

To find the loss in kinetic energy (KE), we first calculate the initial and final kinetic energies.

Initial kinetic energy: $KE_i = rac{1}{2} m_A u_A^2 + rac{1}{2} m_B u_B^2$ Substituting the known values: $KE_i = rac{1}{2}(6)(2^2) + rac{1}{2}(4)(3^2)$ $KE_i = rac{1}{2}(6)(4) + rac{1}{2}(4)(9) = 12 + 18 = 30 ext{ J}$

Final kinetic energy: $KE_f = rac{1}{2} m_A v_A^2 + rac{1}{2} m_B v_B^2$ Substituting the final speeds: $KE_f = rac{1}{2}(6)( rac{6}{5})^2 + rac{1}{2}(4)(- rac{9}{5})^2$ $KE_f = rac{1}{2}(6)( rac{36}{25}) + rac{1}{2}(4)( rac{81}{25})$ $KE_f = rac{108}{25} + rac{162}{25} = rac{270}{25} = 10.8 ext{ J}$

Loss in kinetic energy: $ext{Loss} = KE_i - KE_f = 30 - 10.8 = 19.2 ext{ J}$

Step 3

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

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Answer

Impulse is calculated using the change in momentum: $I = m_A (v_A - u_A)$ Substituting the values we found: $I = 6 imes ( rac{6}{5} - 2)$ Calculating: $I = 6 imes ( rac{6}{5} - rac{10}{5}) = 6 imes (- rac{4}{5}) = - rac{24}{5} = -4.8 ext{ Ns}$

The magnitude of the impulse is: $|I| = 4.8 ext{ Ns}$

Step 4

(i) the speed of the ball when it hits the floor

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Answer

To determine the speed of the ball just before it hits the floor, we can use the equations of motion under constant acceleration.

Using: $v^2 = u^2 + 2as$ Where:

$u = 2 ext{ m/s}$ (initial speed)
$a = 10 ext{ m/s}^2$ (acceleration due to gravity)
$s = 3 ext{ m}$ (height dropped)

Substituting these values gives: $v^2 = (2)^2 + 2(10)(3)$ $v^2 = 4 + 60 = 64$

Thus: $v = ext{sqrt}(64) = 8 ext{ m/s}$ . The speed of the ball when it hits the floor is 8 m/s.

Step 5

(ii) the value of e.

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Answer

To find the coefficient of restitution $e$ , we can use the formula: $e = rac{v_{ ext{rebound}}}{v_{ ext{impact}}}$

The rebound height is given as 1.8 m, so we need to calculate the speed after the rebound. Using the equation: $v_{ ext{rebound}} = ext{sqrt}(2gh)$ Where $h = 1.8 ext{ m}$ : $v_{ ext{rebound}} = ext{sqrt}(2 imes 10 imes 1.8) = ext{sqrt}(36) = 6 ext{ m/s}$

Now substituting back to find $e$ : $e = rac{v_{ ext{rebound}}}{v_{ ext{impact}}} = rac{6}{8} = rac{3}{4}$ Therefore, the value of $e$ is $rac{3}{4}$ .

a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2016

Worked Solution & Example Answer:a) A smooth sphere A, of mass 6 kg, collides directly with another smooth sphere B, of mass 4 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2016

(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.

(i) the speed of the ball when it hits the floor

(ii) the value of e.

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