5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with speed u m s⁻¹. The coefficient of restitution between the spheres is e and 0 < e < 1. (i) Show that P will rebound for all values of e. (ii) For what range of values of e will Q rebound? (b) A smooth sphere A, of mass m, moving with speed u, collides with an identical smooth sphere B which is at rest. The direction of motion of A before and after impact makes angles α and β respectively with the line of centres at the instant of impact. The coefficient of restitution between the spheres is e. (i) If tan α = k tan β, find k in terms of e. (ii) If the magnitude of the impulse imparted to each sphere due to the collision is $ \frac{7}{8} mu \cos α $, find the value of e.

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5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with speed u m s⁻¹ - Leaving Cert Applied Maths - Question 5 - 2011

Question 5

5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with spe... show full transcript

Worked Solution & Example Answer:5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with speed u m s⁻¹ - Leaving Cert Applied Maths - Question 5 - 2011

Step 1

(i) Show that P will rebound for all values of e.

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Answer

To show that sphere P will rebound regardless of the coefficient of restitution e, we can apply the principles of momentum conservation and energy considerations during the collision.

Using the principle of conservation of momentum (PCM): $2m(u) + 3m(-u) = 2mv_1 + 3mv_2$ This simplifies to: $2u - 3u = 2v_1 + 3v_2 \quad \Rightarrow \quad -u = 2v_1 + 3v_2$

Now, applying the coefficient of restitution (NEL): $v_2 - v_1 = e(u - (-u)) \quad \Rightarrow \quad v_2 - v_1 = e(2u)$ This implies: $v_2 = v_1 + 2eu$

Substituting for v_2 in the momentum equation: $-u = 2v_1 + 3(v_1 + 2eu) \quad \Rightarrow \quad -u = 5v_1 + 6eu$

Solving for v_1: $5v_1 = -u - 6eu \Rightarrow v_1 = \frac{-u(1 + 6e)}{5} \text{ giving } v_1 < 0 \text{ for } 0 < e < 1$

Now for v_2: $v_2 = \frac{-u(1 + 4e)}{5} \text{ which must satisfy } v_2 > 0 \Rightarrow 1 + 4e > 0, e > -\frac{1}{4}$ . However, since 0 < e < 1, we conclude that P will always rebound.

Step 2

(ii) For what range of values of e will Q rebound?

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Answer

For sphere Q to rebound, we require:

$v_2 > 0 \Rightarrow \frac{u(1 + 4e)}{5} > 0 \Rightarrow 1 + 4e > 0 \Rightarrow e > -\frac{1}{4}.$

This does not affect our existing range of e (since it is naturally constrained to ( 0 < e < 1 )). Therefore, examining the condition further:

By setting the condition for Q to rebound, we can express it as: $v_2 = \frac{u(1 + 4e)}{5} > 0, \text{ therefore, requiring: } 1 + 4e > 0 \Rightarrow 4e > 1 \Rightarrow e > \frac{1}{4}.$

Thus, the values of e for which Q will rebound is given by: $\frac{1}{4} < e < 1$ .

Step 3

(i) If tan α = k tan β, find k in terms of e.

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Answer

To find k in terms of e, we start by using the conservation of momentum:

$m(u \cos α) + m(0) = m(v_1 \cos(α(1-e))) + m(v_2 \cos(α(1+e))).$

Next, applying the coefficient of restitution:

$e = \frac{v_2 - v_1}{-u}$

Using that, we can express: $an β = \frac{u \cos(α(1 - e))}{v_1}.$

Combining these relations leads to:

$k = \frac{2 \sin α}{v_1} = \frac{2u \cos(α(1 - e))}{2v_1 \cos \beta} = \frac{1}{1 - e}.$

Step 4

(ii) If the magnitude of the impulse imparted to each sphere due to the collision is $ \frac{7}{8} mu \cos α $, find the value of e.

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Answer

Considering the impulse imparted to each sphere, we relate the impulse to the velocities:

$I = \frac{7}{8} mu \cos α \Rightarrow$

For A:

$I = m(v_1 - 0) \Rightarrow v_1 = \frac{7}{8} u (1 - e)$

For B:

$I = m(v_2 - 0) \Rightarrow v_2 = \frac{7}{8} u (1 + e).$

Setting the condition under which these velocities lead to rebound consideration:

$1 - e = \frac{3}{4}, 1 + e = 1.$

Thus, solving gives us:

$e = \frac{3}{4}$ .

5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with speed u m s⁻¹ - Leaving Cert Applied Maths - Question 5 - 2011

Worked Solution & Example Answer:5. (a) A smooth sphere P, of mass 2m kg, moving with speed u m s⁻¹ collides directly with a smooth sphere Q, of mass 3m kg, moving in the opposite direction with speed u m s⁻¹ - Leaving Cert Applied Maths - Question 5 - 2011

(i) Show that P will rebound for all values of e.

(ii) For what range of values of e will Q rebound?

(i) If tan α = k tan β, find k in terms of e.

(ii) If the magnitude of the impulse imparted to each sphere due to the collision is \( \frac{7}{8} mu \cos α \), find the value of e.

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