Question

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.

Answer 1

Josh reaches at
`(h)/(9)` height.

(g) is correct option

Step-by-step 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=(1)/(2)mv^2`

`v=\frac{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)=(√(2gh))/(3)`

We need to calculate the height

Using conservation of energy

`(1)/(2)* m* v^2=mgH`

Put the value into the formula

`(1)/(2)* 3m*((2gh)/(9))=3mgH`

`H=(h)/(9)`

Hence, Josh reaches at
`(h)/(9)` height.