Question

a rope stretched between two boats at rest .a sailor in first board pull the rope with the constant force of 100 Newton first boat with the sailor has a mass of 250 kg where as the mass of the second boat is double the mass of first one and if the initial distance between the boat is 30 then find the time taken for two boat to meet each other(neglect water resistance between boats and water)

Answer 1

Here, the force of 100 N acts on the boat.

Therefore, 250a1 = 100 = a1 = 0.4 m/s2

The relative acceleration value, a = a1 – (-a2) = 0.4 + 0.2 = 0.6 m/s2

Therefore, using the formula S=ut+1/2 at2, we can determine t to be equal to 18.3 s.

Answer 2

Force is the product of mass and acceleration.

Here for the first case the force of 100 N acts on the boat.

We know F=MA1

F=force

M=mass

A1= acceleration (first case)

A= F/M

Or, A= 0.4

Second case

F=MA2 where A2 is acceleration in second case.

A2=F/M

Or, A2= 100/500

Or, A2= 0.2

The acceleration value, A= A1 – (-A2) = 0.4 + 0.2 = 0.6 m/s2

Therefore, using the formula S= UT+1/2AT^2

or, 30=0+1/2×0.6T^2

Or, T^2 = 100

T=10.

So the time required is 10 seconds.