A block of mass m 2.20 kg slides down an incline which is 3.60 m high. At the bo
ID: 1367048 • Letter: A
Question
A block of mass m 2.20 kg slides down an incline which is 3.60 m high. At the bottom, it strikes block mass M 7.00 kg which at rest on a horizontal surface, as 3.60 m shown (assume a smooth transition to the bottom of the incline). If the collision is elastic, and friction can of mass m 2.20 kg just before its strikes the block be ignored, determine (a) the speed of the block of mass M-7.00 kg. (b) the speeds of the two blocks after the collision.
USE these formulas { Va^2=(2*g*h)^2 }
and { Ma*Va^2+Mb*Vb=Ma*Va'+Mb*vb' }
and { 1/2Ma*Va^2+1/2Mb*Vb^2=1/2MaVa'2+1/2Mb*Vb'^2 }
Explanation / Answer
let mass of the first body m1=2.20kg
height of inclination h=3.60m
mass of second body m2=7.00kg
the second body is at rest.so u2=0
the speed of first body u12=2gh
=2x9.8x3.60
u1 =8.4m/sec
the speed of the first body after collision v1=(m1-m2/m1+m2)u1
= (2.20-7.00/2.20+7.00)8.4
=-4.382m/sec
the speed of the second body after collision v2=(2m1/m1+m2)u1
=(2x2.20/2.20+7.00)8.4
=4.017m/sec
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