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Salsk061 [2.6K]
3 years ago
5

two bumper cars at an amusement park collide together abd get stuck together. assuming that the system of the two bumper cars is

isolated. what does the conservation of momentum tell us
Physics
1 answer:
Fiesta28 [93]3 years ago
7 0

The answer to this problem is M1v1-m2v2=(m1+m2)v

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A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja
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Answer: v = 0.6 m/s

Explanation: <u>Momentum</u> <u>Conservation</u> <u>Principle</u> states that for a collision between two objects in an isolated system, the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.

Momentum is calculated as Q = m.v

For the piñata problem:

Q_{i}=Q_{f}

m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}

Before the collision, the piñata is not moving, so v_{p}_{i}=0.

After the collision, the stick stops, so v_{s}_{f}=0.

Rearraging, we have:

m_{s}v_{s}_{i}=m_{p}v_{p}_{f}

v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}

Substituting:

v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}

v_{p}_{f}= 0.6

Immediately after being cracked by the stick, the piñata has a swing speed of 0.6 m/s.

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