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3 years ago

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A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

ection at 14.2 cm/s. After the collision, the 13.5 g marble moves to the right at 71 cm/s. Find the velocity of the 27.0 g marble after the collision.

1 answer:

Nataly [62]3 years ago

7 0

Answer: $v_1=28.4 cm/s$

Explanation:

Given

mass of marble $m_1=27 gm$

velocity of marble $u_1=56.8 cm/s \approx 0.568 m/s$

mass of second marble $m_2=13.5 gm$

Velocity of second marble $u_2=14.2 cm/s \approx 0.142 m/s$

After collision 13.5 gm marble moves to the right at i.e. $v_2=71 cm/s$

Conserving momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71$

$1533.6+191.7=27\cdot v_1+958.5$

$27\cdot v_1=766.8$

$v_1=\frac{766.8}{27}=28.4 cm/s$

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<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

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If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

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$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

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Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

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