The initial speed of a body of mass 2.0kg is 5.0ms−1. 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
8.50N s,9.25ms−1
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B
4.25N s,9.25ms−1
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C
4.25N s,4.25ms−1
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D
8.50N s,4.25ms−1
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Solution
The correct option is A8.50N s,9.25ms−1 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 oa′a + area of reactangle aa′b′b + area of trapezium bb′c′c + area of rectangle cc′ed =12×1.5×3+1×3+12(3+2)0.5+2×1 =2.25+3+1.25+2=8.50N s
Impulse = change in momentum (mΔv)
Where Δv is the increase in velocity. ∴Δv=Impulsem=8.502ms−1=4.25ms−1
Final speed = Initial speed + Δv Final speed 00000=5.00ms−1+4.25ms−1=9.25ms−1