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3 years ago

6

A girl, moving at 8.8 m/s on in-line skates, is overtaking a boy moving at 4.3 m/s as they both skate on a straight path. The bo

y tosses a ball backward toward the girl, giving it speed 11.4 m/s relative to him. What is the speed of the ball relative to the girl, who catches it?

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:16.5 m/s

Explanation:

Given

Velocity of boy $(v_b)=4.3 m/s$

velocity of girl $(v_g)=8.8 m/s$

velocity of ball relative to boy $(v_{Bb})=11.4 m/s$ backward

Now We have to find velocity of ball relative to girl

Representing it in vector form

$v_b=4.3\hat{i}$

$v_g=8.8\hat{i}$

$v_{Bb}=-11.4\hat{i}$

and

$v_{Bb}=v_B-v_b$

$v_B=v_{Bb}+v_b$

$v_B=-11.4\hat{i}+4.3\hat{i}=-7.1\hat{i}$

therefore velocity of ball relative to girl

$v_{Bg}=v_B-v_g=-7.1\hat{i}-8.8\hat{i}=-16.5\hat{i}$

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Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

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Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

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Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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