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3 years ago

A 5 kg ball travels at a velocity of 10 m/s, what is the momentum of the ball

Physics

2 answers:

Sholpan [36]3 years ago

8 0

Answer:

50 kg·m/s.

Explanation:

Momentum can be calculated using p = mv. (Let p represent momentum, m represent mass, and v represent velocity.)

Plugging our terms in, we have p = 5 kg * 10 m/s. This gives us p = 50 kg·m/s as an answer.

Zina [86]3 years ago

4 0

Answer= 59kg•m/s

you do 5kg x 10m/s

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A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at

Lelechka [254]

Answer:

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Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

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We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

$\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

Rearranging , and simplifying common terms, we have:

$m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

$v_{10} + v_{1f} = v_{2f}$

Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

$m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$

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Lelechka [254]

Answer:B

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