Question

Question #91871

Answer 1

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(a) The given slope is `4:10`
i.e., `tanalpha=4/10=0.4`
`=>alpha=tan^-1 0.4=21.8^@`
When the car rolls down `l=12m`,
`Delta h=lsinalpha`
we get
`Delta h=12sin21.8` .......(1)
Loss of Potential energy of the automobile`=mgDeltah`
This gets converted in kinetic energy of the automobile`=1/2mv^2`
where `v` is the velocity of the rolling automobile just before it hits the parked automobile.

Equating the two we get
`1/2mv^2=mgDeltah`
`v=sqrt(2gDeltah)` .....(2)

In the given elastic collision both momentum and kinetic energy are conserved. Assuming that both automobiles move in the same direction down hill we get from momentum
`"Initial Momentum"="Final momentum"`
`1550xxv+850xx0=1550xxv_1+850xxv_2`
where `v_1and v_2` are velocities of the two automobiles after collision.
`=>1550sqrt(2gDeltah)=1550xxv_1+850xxv_2` ........(3)
from Kinetic energy we get
`mgDeltah=1/2xx1550v_1^2+1/2xx850v_2^2` .......(4)
Solve equations (3) and (4) for `v_1and v_2`