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lawyer [7]
3 years ago
11

Part A

Physics
1 answer:
irina [24]3 years ago
3 0

Answer:

v' = -18 m/s

Explanation:

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

       p_{o} = p_{f} (1)

  • The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

       m_{b} * v_{b} -M_{c}*V_{c}  = m_{b} * 18 m/s + (-M_{c}* 20 m/s)  (2)

  • The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
  • So, we can write again (3) as follows:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (4)

  • Replacing (2) and (4) in (1), we get:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (5)

  • Simplifying, and rearranging, we can solve for v'b, as follows:
  • v'_{b} = -18 m/s (6), which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had  inverted its speed after the collision.
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The right answer is:

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The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

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On substituting the values, we get

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⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

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⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

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If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.

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