At the moment $t = 0$, a time dependent force $F = at$ (where $a$ is constant equal to 1 N s$^{-1}$) is applied on a body of mass 1 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 (acceleration due to gravity = 10 m s$^{-2}$)
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When a body leaves a surface, the normal force becomes zero. Use this condition to find the time of departure, then compute the velocity components at that instant.
The body of mass 1 kg is on a smooth horizontal plane with a vertical wall, and a force $F = at$ (where $a = 1 \, \text{N s}^{-1}$) is applied at an angle of $45^\circ$ to the horizontal. We need to find the velocity of the body when it leaves the plane, with $g = 10 \, \text{m s}^{-2}$.
1. Forces and acceleration:
- The force $F = t$ (since $a = 1 \, \text{N s}^{-1}$), so at time $t$, $F = t$.
- Components of the force:
- Horizontal (along the plane, towards the wall): $F_x = F \cos 45^\circ = t \cdot \frac{\sqrt{2}}{2} = \frac{t \sqrt{2}}{2}$.
- Vertical (upward): $F_y = F \sin 45^\circ = t \cdot \frac{\sqrt{2}}{2} = \frac{t \sqrt{2}}{2}$.
- Normal force $N$ from the plane (upward) balances the weight and the vertical component of $F$ until the body leaves the plane:
\[
\begin{align}
N + F_y = mg \implies N + \frac{t \sqrt{2}}{2} = 1 \cdot 10 \implies N = 10 - \frac{t \sqrt{2}}{2}
\]
- The body leaves the plane when $N = 0$:
\[
\begin{align}
10 - \frac{t \sqrt{2}}{2} = 0 \implies \frac{t \sqrt{2}}{2} = 10 \implies t \sqrt{2} = 20 \implies t = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{s}
\]
2. Horizontal motion:
- The horizontal acceleration (since the plane is smooth, only $F_x$ contributes):
\[
\begin{align}
a_x = \frac{F_x}{m} = \frac{\frac{t \sqrt{2}}{2}}{1} = \frac{t \sqrt{2}}{2}
\]
- Acceleration $a_x = \frac{dv_x}{dt} = \frac{t \sqrt{2}}{2}$. Integrate to find velocity:
\[
\begin{align}
v_x = \int_0^t \frac{t' \sqrt{2}}{2} \, dt' = \frac{\sqrt{2}}{2} \int_0^t t' \, dt' = \frac{\sqrt{2}}{2} \left[\frac{(t')^2}{2}\right]_0^t = \frac{\sqrt{2}}{2} \cdot \frac{t^2}{2} = \frac{\sqrt{2} t^2}{4}
\]
- At $t = 10\sqrt{2}$:
\[
\begin{align}
v_x = \frac{\sqrt{2} (10\sqrt{2})^2}{4} = \frac{\sqrt{2} (100 \cdot 2)}{4} = \frac{\sqrt{2} \cdot 200}{4} = 50 \sqrt{2} \, \text{m/s}
\]
3. Vertical motion:
- Vertical acceleration when the body is still on the plane is zero (since $N$ balances forces). Once $N = 0$, the body leaves the plane, and its vertical velocity at that moment is 0 (it just starts to lift off).
4. Velocity when it leaves the plane:
- The velocity has only a horizontal component at the moment it leaves: $v = v_x = 50\sqrt{2} \, \text{m/s}$.
- This matches option (2).
Thus, the correct answer is (2).
