A particle of mass 7 , kg moving at 5 , m/s i | vedclass

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(D) Given: Mass $m = 7 \, kg$,initial velocity $u = 5 \, m/s$. Initial momentum $p_i = m 	imes u = 7 	imes 5 = 35 \, kg \cdot m/s$.Since the force i

Vedclass · Accepted Answer

all the above options are correct

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(D) Given: Mass $m = 7 \, kg$,initial velocity $u = 5 \, m/s$. Initial momentum $p_i = m 	imes u = 7 	imes 5 = 35 \, kg \cdot m/s$.Since the force is opposite to the direction of motion,we take the force as negative: $F = -F_{graph}$.Impulse $J = \int F \, dt = -(	ext{Area under the } F-t 	ext{ graph})$.Area $= \frac{1}{2} 	imes (10 + 4) 	imes 10 = \frac{1}{2} 	imes 14 	imes 10 = 70 \, N \cdot s$.So,change in momentum $\Delta p = -70 \, kg \cdot m/s$.Final momentum $p_f = p_i + \Delta p = 35 - 70 = -35 \, kg \cdot m/s$.Final velocity $v = \frac{p_f}{m} = \frac{-35}{7} = -5 \, m/s$.Since $v = -5 \, m/s$,the speed is $|v| = 5 \, m/s$ (Option $A$ is correct).The negative sign indicates the direction of motion is reversed (Option $B$ is correct).At $t = 5 \, s$,the impulse is $\int_0^5 F \, dt = 	ext{Area of trapezoid from } 0 	ext{ to } 3 + 	ext{Area of rectangle from } 3 	ext{ to } 5 = \frac{1}{2} 	imes 3 	imes 10 + 2 	imes 10 = 15 + 20 = 35 \, N \cdot s$.Momentum at $t = 5 \, s$ is $p(5) = p_i - 35 = 35 - 35 = 0$. Thus,the particle is momentarily at rest (Option $C$ is correct).Therefore,all options are correct.

$A$ particle of mass $7 \, kg$ moving at $5 \, m/s$ is acted upon by a variable force opposite to its initial direction of motion. The variation of force $F$ with time $t$ is shown in the graph.

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