Question #91871

1 Answer
Mar 26, 2017

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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#