A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table. Spheres A and B are moving towards each other with a speed of 5 m s⁻¹. The coefficient of restitution for the collision is rac{2}{3}. 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. (b) 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 3.15 metres. The ball strikes the ceiling with a vertical speed of 12 m s⁻¹. The coefficient of restitution for all collisions between the ball and the ceiling and between the ball and the floor is rac{2}{3}. (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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A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2017

Question 5

A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table. Spheres A and B are moving towards each other wi... show full transcript

Worked Solution & Example Answer:A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2017

Step 1

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

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Answer

To find the speeds of spheres A and B immediately after the collision, we can use the conservation of momentum and the coefficient of restitution. The initial momentum before the collision is:

$ext{Momentum}_{initial} = m_A v_A + m_B v_B = (1 ext{ kg} imes 5 ext{ m/s}) + (3 ext{ kg} imes (-5 ext{ m/s})) = 5 - 15 = -10 ext{ kg m/s}$

Let the final velocities be $v_A'$ for sphere A and $v_B'$ for sphere B. According to the conservation of momentum:

$1 v_A' + 3 v_B' = -10$

The coefficient of restitution is defined as:

$e = rac{v_B' - v_A'}{v_A - v_B}$

Here, $v_A = 5$ m/s and $v_B = -5$ m/s, hence:

$e = rac{v_B' - v_A'}{5 - (-5)} = rac{v_B' - v_A'}{10}$

Given the coefficient of restitution is $rac{2}{3}$ :

$rac{2}{3} = rac{v_B' - v_A'}{10}$

From this, we can deduce:

$v_B' - v_A' = rac{20}{3}$ .

Now we have two equations:

$v_A' + 3v_B' = -10$
$v_B' - v_A' = rac{20}{3}$

Solving these simultaneously will give us the values for $v_A'$ and $v_B'$ .

Step 2

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

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Answer

The kinetic energy before the collision can be calculated as:

$KE_{initial} = rac{1}{2} m_A v_A^2 + rac{1}{2} m_B v_B^2 = rac{1}{2}(1)(5^2) + rac{1}{2}(3)(-5^2) = rac{1}{2}(1)(25) + rac{1}{2}(3)(25) = 12.5 + 37.5 = 50 ext{ J}$

After computing the final velocities $v_A'$ and $v_B'$ , we can find:

$KE_{final} = rac{1}{2} m_A (v_A')^2 + rac{1}{2} m_B (v_B')^2$

The kinetic energy lost due to the collision is:

$ext{Loss of KE} = KE_{initial} - KE_{final}$

Step 3

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

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Answer

The impulse imparted to object B can be calculated using the change in momentum:

$J = ext{Impulse} = m_B (v_B' - v_B)$

Substituting the mass of sphere B and initial and final velocities, we can calculate the magnitude of the impulse.

Step 4

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

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Answer

The speed after striking the ceiling can be found using the coefficient of restitution:

$e = rac{v_{ceiling} - v'}{v_{ceiling}}$

Substituting known values:

$rac{2}{3} = rac{12 - v'}{12}$

Solving for $v'$ gives:

$v' = 12 imes rac{2}{3} = 8 ext{ m/s}$ .

Step 5

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

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Answer

When the ball hits the floor, we again apply the coefficient of restitution:

Using similar calculations:

$v_{floor} = e imes v'_{ceiling} = rac{2}{3} imes 8 = rac{16}{3} ext{ m/s}$

From this speed, we analyze the travel height:

Using $h = rac{v^2}{2g}$ , we find the height reached, and check it against the room height of 3.15 m. Since it reaches above this height, we conclude the ball does strike the ceiling again.

A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2017

Worked Solution & Example Answer:A smooth sphere A, of mass 1 kg, collides with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2017

(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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