Question 13: Laws of Motion

Question

The initial speed of a body of mass 2.0 kg is 5 m/s. A force acts for 4 s in the direction of motion of the body, as shown in force-time graph. The impulse of force is

Accepted Answer

Impulse of a force = Area between the force-time graph
Hence, impulse of force = area of $$\Delta$$OA'A + area of rectangle A'B'BA + area of trapezium BB'C'C + area of rectangle CC'ED
$$ = {1 \over 2} 	imes 1.5 	imes 3 + 1 	imes 3 + {1 \over 2}(3 + 2)(3 - 2.5) + 2 	imes 1$$
$$ = 2.25 + 3 + 1.25 + 2$$
$$ = 8.50$$ Ns

Answer

8.50 Ns

Answer

8 Ns

Answer

5.5 Ns

Answer

6 Ns

Laws of Motion

The initial speed of a body of mass 2.0 kg is 5 m/s. A force acts for 4 s in the direction of motion of the body, as shown in force-time graph. The impulse of force is

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