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aksik [14]
3 years ago
8

Later that day, Martin does the same thing with Josh, who is 2 times heavier than Martin. If the collision is totally inelastic,

what height does Josh reach?
a. h/25
b. h/16
c. h/8
d. h/4
e. h/2
f. h
g. None of the above.
Physics
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

Josh reaches at  \dfrac{h}{9}  height.

(g) is correct option

Explanation:

Given that,

Mass of martin = m

Mass of josh m'= 2m

Martin and Paul are skateboarding in a large semicircular halfpipe. Martin starts out from rest at a height h and collides with Paul standing at the bottom. Martin and Paul have about the same mass. After the collision,

We need to calculate the speed before the collision

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

v=\dfrac{2gh}

We need to calculate the final speed after collision

Using conservation of momentum

mv_{i}=(m+2m)v_{f}

Put the value into the formula

v_{f}=\dfrac{\sqrt{2gh}}{3}

We need to calculate the height

Using conservation of energy

\dfrac{1}{2}\times m\times v^2=mgH

Put the value into the formula

\dfrac{1}{2}\times 3m\times(\dfrac{2gh}{9})=3mgH

H=\dfrac{h}{9}

Hence, Josh reaches at \dfrac{h}{9} height.

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