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2 years ago

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Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc

ity of car B after the collision?

1 answer:

Nuetrik [128]2 years ago

4 0

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

$v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2$

The data given are;

m1 = 281kg
u1 = 2.82m/s
m2 = 209kg
u2 = -1.72m/s
v1 = ?

Let's substitute the values into the equation.

$v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s$

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

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Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

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Explanation:

Given:

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$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

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On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

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<u>Now the net effective speed of stream sweeping the swimmer:</u>

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$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

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<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

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Now let the distance swam in this direction be d'.

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$d'=\frac{500}{\cos63.39}$

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$\Delta s'=\sqrt{1116.344^2-500^2}$

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3)

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Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

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