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A small smooth sphere A, of mass 1.5 kg, moving with speed 6 m s−1, collides directly with a small smooth sphere B, of mass m kg, which is at rest - Leaving Cert Applied Maths - Question 5 - 2017
Question 5
A small smooth sphere A, of mass 1.5 kg, moving with speed 6 m s−1, collides directly with a small smooth sphere B, of mass m kg, which is at rest. After the collisi... show full transcript
Worked Solution & Example Answer:A small smooth sphere A, of mass 1.5 kg, moving with speed 6 m s−1, collides directly with a small smooth sphere B, of mass m kg, which is at rest - Leaving Cert Applied Maths - Question 5 - 2017
Step 1
(i) the value of v
Answer
To find the value of v, we can use the principle of conservation of momentum and the equation for kinetic energy.
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Conservation of Momentum: The total initial momentum before the collision equals the total momentum after the collision.
[ 1.5 \times 6 = 1.5v + m(2v) ]
Solving for the mass of sphere B: [ 9 = 1.5v + 2mv ] [ 9 = v(1.5 + 2m) ] [ v = \frac{9}{1.5 + 2m} \tag{1} ]
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Kinetic Energy Lost: 80% of the kinetic energy lost by A is transferred to B.
Initial kinetic energy of A: [ KE_A = \frac{1}{2}(1.5)(6^2) = 54 \text{ J} ]
After collision, kinetic energies of A and B are:
[ KE_{A, final} = \frac{1}{2}(1.5)v^2 \text{ J} ] and [ KE_{B, final} = \frac{1}{2}m(2v)^2 = 2mv^2 \text{ J} ]
The total kinetic energy lost: [ \Delta KE = 54 - (\frac{1}{2}(1.5)v^2 + 2mv^2) ]
Setting this equal to 80% of the initial: [ 0.8 \times 54 = 54 - (\frac{1}{2}(1.5)v^2 + 2mv^2) ]
Rearranging gives: [ (\frac{1}{2}(1.5)v^2 + 2mv^2) = 10.8 ]
By substituting equation (1) into the energy equation, we can solve for v:
After calculating, we find: [ v = \frac{12}{7} \text{ m/s} ]
Step 2
(ii) the value of e
Answer
Now we need to determine the coefficient of restitution e.
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Coefficient of Restitution Definition: The coefficient of restitution e is defined as:
[ e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} ]
For this scenario:
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Relative speed of approach: The relative speed of sphere A before collision is: ( 6 , ext{m/s} ).
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Relative speed of separation: After the collision, sphere A moves with speed v and sphere B moves with speed 2v. The relative speed of separation is then: [ |v - (2v)| = | -v | = v ]
- Calculating e:
Substituting these values back into the e formula: [ e = \frac{v}{6} ]
Substituting our result from (i): [ e = \frac{\frac{12}{7}}{6} = \frac{2}{7} ]
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Step 3
(i) Find, in terms of v and α, the speed of P before the collision.
Answer
To find the speed of P before the collision, we apply the principle of conservation of momentum (PCM) and the laws of motion.
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Conservation of Momentum: The total momentum before the collision is equal to the total momentum after the collision:
[ 3m(u \cos \alpha + 0) + 7m(0 + 0) = 3m(u' \cos 0 + v \cos \alpha) + 7m(u \sin \alpha) ]
Simplifying: [ 3u \cos \alpha = 3u' + 7v \cos \alpha ]
- Using Coefficient of Restitution: Given e = ( \frac{2}{7} ), we have:
[ v' - 0 = e(0 - u \cos \alpha) \implies u' = \frac{2}{7}u \cos \alpha ]
Substituting this back: [ 3u \cos \alpha = 3(\frac{2}{7}u \cos \alpha) + 7v \cos \alpha ]
You will find that: [ u = \frac{70v}{27 \cos \alpha} ]
Step 4
(ii) If α = 30° find the angle through which the direction of motion of P is deflected as a result of the collision.
Answer
To find the deflection angle, we employ trigonometry based on the geometry post-collision.
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Calculating speeds: Given α = 30°: Substitute α into the equation we derived:
[ u = \frac{70v}{27 \cos(30)} = \frac{70v}{27 \cdot \frac{\sqrt{3}}{2}} ]
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Using tangent to find the angle of deflection: Take the tangent of the angles involved:
[ \tan(\beta) = \frac{v \sin(30)}{u \cos(30)} ]
Substitute u and simplify: [ \tan(\beta) = \frac{10v}{u \sqrt{3}} = \frac{10v}{(70v/27 \cdot \frac{\sqrt{3}}{2}) \sqrt{3}} ]
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Final calculations: Solve for β: [ \beta = \tan^{-1}(\frac{10 \cdot 2}{70}) ] yielding an angle of approximately 50.17°. The final angle through which the direction of motion of P is deflected is: [ \beta ext{ = 50.17°} ]