Motion In A Plane Chapter-Wise Test 3

Frequency \( f = \frac{n}{t} = \frac{4}{8} = 0.5 \, \text{Hz} \).

Angular speed \( \omega = 2\pi f = 2\pi \times 0.5 = \pi \, \text{rad/s} \).

Centripetal acceleration \( a_c = \omega^2 R = (\pi)^2 \times 10 \approx 9.87 \times 10 \approx 98.7 \, \text{m/s}^2 \).

If the centripetal force disappears, the object will move tangentially to the circle in a straight line due to inertia (Newton’s first law). Without the force, there is no acceleration to keep it in the circular path.

Maximum height \( h_m = \frac{(v_0 \sin \theta_0)^2}{2g} \).

Given: \( v_0 = 28 \, \text{m/s}, \theta_0 = 45^\circ, \sin 45^\circ = \frac{1}{\sqrt{2}}, g = 9.8 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 28 \times \frac{1}{\sqrt{2}} = 14\sqrt{2} \approx 19.8 \, \text{m/s} \).

\( h_m = \frac{(19.8)^2}{2 \times 9.8} = \frac{392.04}{19.6} \approx 20 \, \text{m} \).

Vertical motion: \( y = \frac{1}{2} g t^2 \), where \( y = -45 \, \text{m} \).

\( -45 = \frac{1}{2} \times 10 \times t^2 \Rightarrow -45 = 5 t^2 \Rightarrow t^2 = 9 \Rightarrow t = 3 \, \text{s} \).

Velocity components: \( v_x = 8 \, \text{m/s}, v_y = 6 \, \text{m/s} \).

Magnitude \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{m/s} \).

At maximum height, speed = \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 50 \, \text{m/s}, \cos 53^\circ = 0.6 \).

\( v_x = 50 \times 0.6 = 30 \, \text{m/s} \).

Horizontal velocity \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 22 \, \text{m/s}, \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

\( v_x = 22 \times 0.866 \approx 19.05 \, \text{m/s} \).

Horizontal velocity v_x = v₀ cos θ₀.

Given: v₀ = 30 m/s, cos 30° = √3/2.

v_x = 30 × (√3/2) = 15 √3 ≈ 25.98 m/s.

Time to maximum height \( t_m = \frac{v_0 \sin \theta_0}{g} \).

Given: \( v_0 = 16 \, \text{m/s}, \sin 60^\circ = 0.866, g = 10 \, \text{m/s}^2 \).

\( t_m = \frac{16 \times 0.866}{10} \approx \frac{13.856}{10} \approx 1.39 \, \text{s} \).

The centripetal force is always perpendicular to the velocity (tangential) in uniform circular motion. Since work is the dot product of force and displacement, and the displacement is along the velocity, the work done is zero (force and displacement are perpendicular).