Question

The potential energy of a particle of mass 5 kg moving in xy-plane is given as $U = \left( {7x + 24y} \right)\,{\text{joule}}$, x and y being in meter. Initially at $t = 0$, the particle is at the origin $\left( {0,0} \right)$ moving with a velocity of $\left( {8.6\hat i + 23.2\hat j} \right)\,{\text{m/s}}$. Then
A. The velocity of the particle at $t = 4\,{\text{s}}$, is $5\,{\text{m}}{{\text{s}}^{ - 1}}$.
B. The acceleration of the particle is $5\,{\text{m}}{{\text{s}}^{ - 2}}$.
C. The direction of motion of the particle initially (at $t = 0$) is at right angles to the direction of acceleration.
D. The path of the particle is a circle.

Answer 1

Calculate the acceleration of the particle using Newton’s second law of motion. Express the kinematic equation for final velocity in terms of its horizontal components and then in terms of vertical components. Use $v = \sqrt {v_x^2 + v_y^2}$ to determine the final velocity. Do the same for acceleration. If the direction of motion of the particle initially (at $t = 0$) is at right angles to the direction of acceleration, then the dot product of acceleration and initial velocity should be zero.

Formula used:
The relation between force and potential energy is,
$F = - \dfrac{{\partial U}}{{\partial r}}$
Here, U is the potential energy and r is the distance.
Kinematic equation, $v = u + at$, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step solution:
To answer this question, we have to verify every given option. We have given the potential energy of the particle $U = \left( {7x + 24y} \right)\,{\text{joule}}$ and initial velocity of the particle $\vec u = \left( {8.6\hat i + 23.2\hat j} \right)\,{\text{m/s}}$.

To determine the acceleration of the particle, we have to find out the force acting on the particle. We have,
$F = - \dfrac{{\partial U}}{{\partial r}}$
$\Rightarrow F = - \left( {\dfrac{{\partial U}}{{\partial x}} + \dfrac{{\partial U}}{{\partial y}}} \right)$

Substituting $U = \left( {7x + 24y} \right)\,{\text{joule}}$ in the above equation, we get,
$\Rightarrow F = - \left( {\dfrac{\partial }{{\partial x}}\left( {7x + 24y} \right) + \dfrac{\partial }{{\partial y}}\left( {7x + 24y} \right)} \right)$
$\Rightarrow \vec F = - \left( {7\hat i + 24\hat j} \right)\,{\text{N}}$

The acceleration of the particle is,
$\vec a = \dfrac{{\vec F}}{m}$ , where m is the mass of the particle.

Substituting $\vec F = - \left( {7\hat i + 24\hat j} \right)\,{\text{N}}$ and $m = 5\,kg$ in the above equation, we get,
$\vec a = \dfrac{{ - \left( {7\hat i + 24\hat j} \right)\,}}{5}$
$\Rightarrow \vec a = - \dfrac{7}{5}\hat i - \dfrac{{24}}{5}\hat j$ …… (1)

From this equation, we can write,
${a_x} = - \dfrac{7}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$ and ${a_y} = - \dfrac{{24}}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$

We have the kinematic equation,
$\vec v = \vec u + \vec at$

Let’s determine the x-component of the final velocity as follows,
${v_x} = {u_x} + {a_x}t$

Substituting ${a_x} = - \dfrac{7}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$ and ${u_x} = 8.6\,{\text{m/s}}$ and 4 s for t in the above equation, we get,
${v_x} = 8.6 + \left( { - \dfrac{7}{5}\,} \right)\left( 4 \right)$
$\Rightarrow {v_x} = 3\,{\text{m}}{{\text{s}}^{ - 1}}$

Let’s determine the y-component of the final velocity as follows,
${v_y} = {u_y} + {a_y}t$

Substituting ${a_y} = - \dfrac{{24}}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$ and ${u_y} = 23.2\,{\text{m/s}}$ in the above equation, we get,
${v_y} = 23.2 - \left( {\dfrac{{24}}{5}} \right)\,\left( 4 \right)$
$\Rightarrow {v_y} = 4\,{\text{m}}{{\text{s}}^{ - 1}}$

Therefore, we can determine the final velocity of the particle as follows,
$v = \sqrt {v_x^2 + v_y^2}$

Substituting ${v_x} = 3\,{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_y} = 4\,{\text{m}}{{\text{s}}^{ - 1}}$in the above equation, we get,
$v = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}}$
$\Rightarrow v = \sqrt {25}$
$\Rightarrow v = 5\,{\text{m}}{{\text{s}}^{ - 1}}$

Therefore, the velocity of the particle is $5\,{\text{m}}{{\text{s}}^{ - 1}}$ after 4 s. Thus, the option (A) is correct.

Let’s determine the acceleration of the particle as follows,
$a = \sqrt {a_x^2 + a_y^2}$
$\Rightarrow a = \sqrt {{{\left( { - \dfrac{7}{5}} \right)}^2} + {{\left( { - \dfrac{{24}}{5}} \right)}^2}}$
$\Rightarrow a = \sqrt {25}$
$\Rightarrow a = 5\,{\text{m}}{{\text{s}}^{ - 2}}$

Therefore, the acceleration of the particle is $5\,{\text{m}}{{\text{s}}^{ - 2}}$. Thus, the option (B) is correct.

If the direction of motion of the particle initially (at $t = 0$) is at right angles to the direction of acceleration, then the dot product of acceleration and initial velocity should be zero.
$\vec a \cdot \vec u = \left( { - \dfrac{7}{5}\hat i - \dfrac{{24}}{5}\hat j} \right) \cdot \left( {8.6\hat i + 23.2\hat j} \right)$
$\Rightarrow \vec a \cdot \vec u = - 12.4 - 18.4$
$\Rightarrow \vec a \cdot \vec u = - 30.8$
$\Rightarrow \vec a \cdot \vec u = 0$

Therefore, the option (C) is incorrect.

Using the kinematic equation, we can express the x-coordinate of the motion of the particle as follows,
$x = {u_x}t + \dfrac{1}{2}{a_x}{t^2}$

Substituting ${a_x} = - \dfrac{7}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$ and ${u_x} = 8.6\,{\text{m/s}}$ in the above equation, we get,
$x = \left( {8.6} \right)t + \dfrac{1}{2}\left( { - \dfrac{7}{5}} \right){t^2}$
$\Rightarrow x = 8.6t - \dfrac{7}{{10}}{t^2}$ ……. (2)

Also, using the kinematic equation, we can express the y-coordinate of the motion of the particle as follows,
$y = {u_y}t + \dfrac{1}{2}{a_y}{t^2}$

Substituting ${a_y} = - \dfrac{{24}}{5}\,{\text{m}}{{\text{s}}^{ - 2}}$ and ${u_y} = 23.2\,{\text{m/s}}$ in the above equation, we get,
$y = 23.2\,t + \dfrac{1}{2}\left( { - \dfrac{{24}}{5}\,} \right){t^2}$
$\Rightarrow y = 23.2\,t - \dfrac{{24}}{{10}}{t^2}$ …… (3)

Now, we know that for a particle to have circular path,
${x^2} + {y^2} = {\text{constant}}$

Substituting equation (2) and (3) in the above equation, we have,
${\left( {8.6t - \dfrac{7}{{10}}{t^2}} \right)^2} + {\left( {23.2\,t - \dfrac{{24}}{{10}}{t^2}} \right)^2} \ne {\text{constant}}$

The left-hand side will never be the constant value. It will have a time term. Therefore, the option (D) is incorrect.

So, the correct answer is option (A) and (B).

Note:
The force is always a negative partial derivative of the potential energy. We use partial derivative when the function depends on more than one variable. The dot product of acceleration and initial velocity is expressed as $\vec a \cdot \vec u = au\sin \theta$. If the dot product of acceleration and initial velocity is zero, then $\sin \theta$ must be zero. This implies that the angle is $90^\circ$.