Question

A particle of mass 2 kg initially at rest starts moving under the force $\vec{F} = 6t^2\hat{i} - 4at\hat{j}$.
If the speed of the particle at $t = 1$ s is $\sqrt{5}$, then the value of constant a is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

This problem is a direct application of the integral form of Newton's second law. Remember that integration introduces a constant, which must be determined from the initial conditions of the problem. Also, distinguish carefully between velocity (a vector) and speed (the magnitude of the velocity vector).

Answer 1

The Correct Option is B

Note:

The force acting on the particle is:
F(t) = 6t² î − 4at ĵ

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Step 1: Use impulse–momentum theorem

According to the impulse–momentum theorem:

Change in momentum = Impulse

mv = ∫ F(t) dt

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Step 2: Apply impulse–momentum separately along x and y directions

Given mass, m = 2 kg
Initial velocity = 0

Along x-direction:

2v_x = ∫ 6t² dt

2v_x = 2t³

v_x = t³

Along y-direction:

2v_y = ∫ (−4at) dt

2v_y = −2at²

v_y = −at²

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Step 3: Write velocity vector at t = 1 s

v_x(1) = 1
v_y(1) = −a

So,

v(1) = î − aĵ

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Step 4: Use given speed condition

Speed at t = 1 s is √5 m/s.

|v| = √(1² + a²)

√(1 + a²) = √5

1 + a² = 5

a² = 4

a = 2

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Final Answer:

The value of the constant is
a = 2