Question

80. Two trolleys of mass m and 3m are connected
by a spring. They are compressed and released,
they moves off in opposite direction and comes
to rest after covering distances s, and s,
respectively. If the frictional force acting
between trolley and surface is same in both
cases then the ratio of distances s,: s, is
1) 1:9 2) 1:3 3) 3:1 4) 9:1​

Answer 1

Answer:

3:1

Explanation:

m1 = m

m2= 3m

law of conservation of linear mumentum

m1v1^2=m2v2^2

mv1^2=3mV2^2

V1^2= 3 V2^2 …......... (1)

now applying  

V1^2= 2aS1

3V2^2= 2a S1

now for s2

V2^2= 2aS2 …......... (2)

by dividing equation 1 / 2  

will get 3/1 =S1/S2

S1:S2 = 3:1