Question

1. The motion of a body of mass 5 kg is shown in the Figure 2.15.
Find (a) its acceleration (b) the force acting on the body (c) change
momentum of bodu 2 s after the start.
(CBSE 2011)​

Answer 1

Answer:

We know that the Newton's second law of motion is F=

Δt

Δp

or change in momentum

or the gain in momentum is Δp=FΔt=Area under the F-t graph

The net area till the end of t=10s is consists of a trapezium with parallel side as 10second and 8−4=4second

and the separation being 20N so area or momentum is

2

sum×separation

=

2

(10+4)second×20N

=140Newtonsecond

Explanation:

We know that the Newton's second law of motion is F=

Δt

Δp

or change in momentum

or the gain in momentum is Δp=FΔt=Area under the F-t graph

The net area till the end of t=10s is consists of a trapezium with parallel side as 10second and 8−4=4second

and the separation being 20N so area or momentum is

2

sum×separation

=

2

(10+4)second×20N

=140Newtonsecond