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bagirrra123 [75]
3 years ago
13

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

aves with a momentum of +13 kg • m/s.
What is the magnitude of the final momentum of the bowling pin if it has a mass of 1.5 kg?
Physics
1 answer:
dangina [55]3 years ago
8 0

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

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Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

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Answer:

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Explanation:

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Therefore it is advisable that she increase the tension of the string.

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A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large
luda_lava [24]
A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large as the droplet?

Answer: In this case the density will remain constant for both droplets. The reason being that volume will not change the density of the material. The only way of changing it is by changing its state. If you increase the volume then the mass will also increase. Leaving the density the same.

I hope it helps, Regards.
6 0
3 years ago
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