Question

At the moment \(t=0\), a time dependent force \[ F=at \] where \(a\) is a constant equal to \(1\,\text{N s}^{-1}\), is applied to a body of mass \(1\,\text{kg}\) resting on a smooth horizontal plane as shown in the figure. If the direction of this force makes an angle \(45^\circ\) with the horizontal, then the velocity of the body at the moment it leaves the plane is \((g=10\,\text{m/s}^2)\):

• A. \(50\,\text{m/s}\)
• B. \(50\sqrt{2}\,\text{m/s}\)
• C. \(100\sqrt{2}\,\text{m/s}\)
• D. \(100\,\text{m/s}\)

Hint:

For a body leaving a horizontal plane, use the condition \[ N=0 \] which means the upward component of applied force becomes equal to the weight.

Answer 1

The Correct Option is B

Step 1: Identify when the body leaves the plane.
The body lies on a smooth horizontal plane. It leaves the plane when the normal reaction from the plane becomes zero.
At that moment, the upward vertical component of force exactly equals the weight of the body.

Step 2: Write the condition for leaving the plane.
The vertical component of the applied force is $F\sin 45^\circ = \frac{F}{\sqrt{2}}$.
At the instant of leaving:
\[ \frac{F}{\sqrt{2}} = mg \] Since $F = at = t$ (with $a = 1\,\text{N\,s}^{-1}$) and $m = 1\,\text{kg}$, $g = 10\,\text{m/s}^2$:
\[ \frac{t_0}{\sqrt{2}} = 1 \times 10 \implies t_0 = 10\sqrt{2}\,\text{s} \]

Step 3: Write the horizontal component of force as a function of time.
The horizontal component is:
\[ F_x = F\cos 45^\circ = \frac{t}{\sqrt{2}} \]

Step 4: Apply the impulse-momentum theorem for the horizontal direction.
The body starts from rest ($v_{x0} = 0$). The horizontal impulse equals the change in horizontal momentum:
\[ mv_x = \int_0^{t_0} F_x\,dt = \int_0^{10\sqrt{2}} \frac{t}{\sqrt{2}}\,dt \]

Step 5: Evaluate the integral.
\[ mv_x = \frac{1}{\sqrt{2}} \left[\frac{t^2}{2}\right]_0^{10\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{(10\sqrt{2})^2}{2} = \frac{1}{\sqrt{2}} \cdot \frac{200}{2} = \frac{100}{\sqrt{2}} \] Since $m = 1\,\text{kg}$:
\[ v_x = \frac{100}{\sqrt{2}} = 50\sqrt{2}\,\text{m/s} \]

Step 6: Find the speed at the moment of leaving.
At the instant of leaving, the normal force is zero, so the net vertical velocity is also being built up. However, the question asks for the speed of the body at the moment it leaves the plane. The horizontal speed at that instant is $50\sqrt{2}\,\text{m/s}$ and there is no horizontal contribution from the vertical at the boundary moment.
\[ \boxed{50\sqrt{2}\,\text{m/s}} \]