(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction. The coefficient of restitution between the spheres is $rac{1}{2}$. (i) Find, in terms of u, the speed of each sphere after the collision. After the collision B hits a smooth vertical wall which is perpendicular to the direction of motion of B. The coefficient of restitution between B and the wall is $rac{1}{2}$. The first collision between the spheres occurred at a distance 2 metres from the wall. The spheres collide again 4 seconds after the first collision between them. (ii) Find the value of u. (b) A smooth sphere P, of mass 2m, collides with a smooth sphere Q, of mass m. The velocity of P is $3u + 4u$ and the velocity of Q is $-4u + 3u$, where $ar{i}$ is along the line of centres at impact. Find (i) In terms of u, the speed of each sphere after the collision (ii) the angle between the directions of P and Q after the collision.

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(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction - Leaving Cert Applied Maths - Question 5 - 2019

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(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction.... show full transcript

Worked Solution & Example Answer:(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction - Leaving Cert Applied Maths - Question 5 - 2019

Step 1

Find, in terms of u, the speed of each sphere after the collision.

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Answer

To find the speeds after the collision, we use the conservation of momentum and the definition of the coefficient of restitution. Let $v_1$ be the speed of sphere A and $v_2$ be the speed of sphere B after the collision.

Using conservation of momentum: $3m imes u + m imes (-u) = 3m imes v_1 + m imes v_2$ This simplifies to: $3u - u = 3v_1 + v_2$ which leads to: $2u = 3v_1 + v_2$

Applying the coefficient of restitution: $e = rac{v_2 - v_1}{u - (-u)}$ Given that $e = rac{1}{2}$ , this can be written as: $rac{1}{2} = rac{v_2 - v_1}{2u}$ This gives us: $v_2 - v_1 = u$

Now, we have a system of equations:

$2 = 3v_1 + v_2$
$v_2 - v_1 = u$

From equation (2), we get:
$v_2 = v_1 + u$

Substituting into equation (1): $2 = 3v_1 + (v_1 + u)$ This gives: $2 = 4v_1 + u$ Rearranging yields:

ightarrow v_1 = rac{2 - u}{4}$$ Then using $v_2 = v_1 + u$: $$v_2 = rac{2 - u}{4} + u = rac{2 - u + 4u}{4} = rac{2 + 3u}{4}$$ Therefore, the speeds after the collision are: $$v_1 = rac{2 - u}{4}$$ $$v_2 = rac{2 + 3u}{4}$$

Step 2

Find the value of u.

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Answer

To find u, we analyze the time and distance traveled. The spheres collide again 4 seconds after their first collision. The distance to the wall is 2 m, and the time for sphere B to reach the wall is given by: $t = rac{8u}{5u} = rac{8}{5} ext{ seconds}$

The remaining time until the next collision is: $4 - rac{8}{5} = rac{20 - 8}{5} = rac{12}{5} ext{ seconds}$

During this time, sphere A travels: $ext{Distance} = v_1 imes ext{Time} = rac{2 - u}{4} imes rac{12}{5}$

Setting up the equation to find u, where the distances covered by both spheres equals 2 m, leads to: $rac{3u}{4} imes rac{12}{5} = 2$

This simplifies to: $u = 1.6 ext{ m/s}$

Step 3

In terms of u, the speed of each sphere after the collision.

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Answer

Using the previous relationships derived from momentum and restitution, substitute u back into the equations for $v_1$ and $v_2$ : $v_1 = rac{2 - u}{4} ext{ and } v_2 = rac{2 + 3u}{4}$ Now re-evaluating based on the u determined: When $u = 1.6$ : $v_1 = rac{2 - 1.6}{4} = 0.10 ext{ m/s}$ $v_2 = rac{2 + 3 imes 1.6}{4} = 1.08 ext{ m/s}$

Step 4

The angle between the directions of P and Q after the collision.

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Answer

To find the angle $heta$ between the velocities of P and Q after the collision: Utilizing the relationship: $tan( heta) = rac{v_{2y}}{v_{2x}}$ for both velocities: Using the velocities:
v1 = $3u + 4u$ and v2 = $-4u + 3u$ , we find values for their components after the collsion: $v_{P} = (u_{2}, v_{2})$ $v_{Q} = (v_{x}, v_{y})$ Substituting gives: tan^{-1} rac{rac{4}{3}} = 14.04 $and$ tan^{-1} rac{rac{(-4)}{-3}} = 104.04 $The angle $ heta$ can then be found using:$ heta = 14.04 + 53.13- (-36.87) = total.$$

(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction - Leaving Cert Applied Maths - Question 5 - 2019

Worked Solution & Example Answer:(a) A small smooth sphere A, of mass 3m moving with speed u, collides directly with a small smooth sphere B, of mass m moving with speed u in the opposite direction - Leaving Cert Applied Maths - Question 5 - 2019

Find, in terms of u, the speed of each sphere after the collision.

Find the value of u.

In terms of u, the speed of each sphere after the collision.

The angle between the directions of P and Q after the collision.

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