A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table. A and B are moving in opposite directions with speeds of 3 m/s and 2 m/s, respectively. The coefficient of restitution for the collision is $rac{1}{2}$. 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 B due to the collision.

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A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2013

Question 5

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A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table. A and B are moving in opposite directio... show full transcript

Worked Solution & Example Answer:A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2013

Step 1

the speed of A and the speed of B after the collision

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Answer

Let the speeds of A and B after the collision be denoted as $v_a$ and $v_b$ respectively.

Using the principle of conservation of linear momentum: $4(3) + 1(-2) = 4v_a + 1v_b$ Thus, $12 - 2 = 4v_a + v_b \\ 10 = 4v_a + v_b \\ (1)$

Using the coefficient of restitution: $e = \frac{v_b - v_a}{u_a - u_b}$ Where:

$u_a = 3 ext{ m/s}$ (speed of A before collision)
$u_b = -2 ext{ m/s}$ (speed of B before collision)

Plugging in the values: $\frac{1}{2} = \frac{v_b - v_a}{3 - (-2)} = \frac{v_b - v_a}{5} \\ v_b - v_a = \frac{5}{2} \\ (2)$

Now, we have two equations:

$10 = 4v_a + v_b$
$v_b - v_a = \frac{5}{2}$

From equation (2), we get: $v_b = v_a + \frac{5}{2} \\ (3)$

Substituting (3) in (1): $10 = 4v_a + (v_a + \frac{5}{2}) \\ 10 = 5v_a + \frac{5}{2} \\ 5v_a = 10 - \frac{5}{2} \\ 5v_a = \frac{20}{2} - \frac{5}{2} \\ 5v_a = \frac{15}{2} \\ v_a = \frac{3}{2} ext{ m/s} \\ v_a = 1.5 ext{ m/s}$

Now substitute $v_a$ back into (3) to find $v_b$ : $v_b = 1.5 + 2.5 = 4 ext{ m/s}$

Thus, the speeds after the collision are:

Speed of A ( $v_a$ ) = 1.5 m/s
Speed of B ( $v_b$ ) = 4 m/s.

Step 2

the loss in kinetic energy due to the collision

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Answer

The initial kinetic energy (KE) before the collision can be calculated as:

$KE_i = \frac{1}{2} m_a u_a^2 + \frac{1}{2} m_b u_b^2$ where:

$m_a = 4 ext{ kg}$ (mass of A)
$u_a = 3 ext{ m/s}$ (initial speed of A)
$m_b = 1 ext{ kg}$ (mass of B)
$u_b = -2 ext{ m/s}$ (initial speed of B)

Calculating: $KE_i = \frac{1}{2} (4)(3^2) + \frac{1}{2} (1)(-2)^2 \\ = \frac{1}{2} (4)(9) + \frac{1}{2} (1)(4) \\ = 18 + 2 = 20 ext{ J}$

The final kinetic energy after the collision is: $KE_f = \frac{1}{2} m_a v_a^2 + \frac{1}{2} m_b v_b^2$ Calculating: $KE_f = \frac{1}{2} (4)(1.5^2) + \frac{1}{2} (1)(4^2) \\ = \frac{1}{2} (4)(2.25) + \frac{1}{2} (1)(16) \\ = 9 + 8 = 12.5 ext{ J}$

The loss in kinetic energy is then: $\Delta KE = KE_i - KE_f = 20 - 12.5 = 7.5 ext{ J}$

Step 3

the magnitude of the impulse imparted to B due to the collision

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Answer

Impulse (I) can be calculated using the formula: $I = m(v_b - u_b)$ where:

$m = 1 ext{ kg}$ (mass of B)
$u_b = -2 ext{ m/s}$ (initial speed of B)
$v_b = 4 ext{ m/s}$ (final speed of B)

Now substituting: $I = 1(4 - (-2)) = 1(4 + 2) = 1(6) = 6 ext{ Ns}$

Thus, the magnitude of the impulse imparted to B is 6 Ns.

A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2013

Worked Solution & Example Answer:A smooth sphere A, of mass 4 kg, collides directly with another smooth sphere B, of mass 1 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2013

the speed of A and the speed of B after the collision

the loss in kinetic energy due to the collision

the magnitude of the impulse imparted to B due to the collision

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