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Question

The initial speed of a body of mass 2.0 kg is 5.0 ms1. A force acts for 4 seconds in the direction of motion of the body. The force-time graph is shown in figure. Calculate the impulse of the force and also the final speed of the body.

A
8.50 N s,9.25 ms1
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B
4.25 N s,9.25 ms1
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C
4.25 N s,4.25 ms1
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D
8.50 N s,4.25 ms1
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Solution

The correct option is A 8.50 N s,9.25 ms1
The magnitude of the impulse of a force is equal to the area between the force-time curve and the time-axis.
Impulse of given force i.e., J= area of oabcde
J= area of triangle oaa + area of reactangle aabb + area of trapezium bbcc + area of rectangle cced
=12×1.5×3+1×3+12(3+2)0.5+2×1
=2.25+3+1.25+2=8.50 N s
Impulse = change in momentum (mΔv)
Where Δv is the increase in velocity.
Δv=Impulsem=8.502 ms1=4.25 ms1
Final speed = Initial speed + Δv
Final speed 00000=5.00 ms1+4.25 ms1=9.25 ms1

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