Motion in a Plane

Q: What is the linear velocity of a person at equator of the earth to its spinning motion ?  (Radius of the earth = 6400 km).

Sol. The earth completes one rotation in 24 hour. Its angular velocity.

$\large \omega = \frac{2\pi N}{t}$

$\large \omega = \frac{2\pi \times 1}{24 \times 60 \times 60} =\frac{\pi}{43,200} rad/s $

The linear velocity,

$\large v = r \omega = 6.4 \times 10^6 \times \frac{\pi}{43,200} $

= 465.5 m/s

Q: When a motor cyclist takes a U- turn in 4s what is the average angular velocity of the motor cyclist.

Sol. When the motor cyclist takes a U – turn, angular displacement, θ = π rad and t = 4s.

The average angular velocity,

ω= θ/t = π /4=0.7855 rad /s

Q: A projectile has the maximum range of 500 m. if the projectile is now thrown up on an inclined plane of 30° with the same speed, what is the distance covered by it along the inclined plane ?

Sol: $\large R_{max} = \frac{u^2}{g}$

$\large 500 = \frac{u^2}{g}$

$\large u^2 = 500 g $

$\large v^2 = u^2 + 2as $

$\large 0 = 500 g + 2(-gsin30^o)s $

s = 500 m

Q: From the top of a tower, two balls are thrown horizontally with velocities u₁ and u₂ in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, find the height or tower.

Sol: Let h = height of the tower

$\large t = \sqrt{\frac{2 h}{g}}$

At time of reaching ground velocities are $\vec{v_1} = u_1 \hat{i} – g t\hat{j}$ and $\vec{v_2} = – u_2 \hat{i} – g t\hat{j}$

As velocities are perpendicular to each other hence ,

$\large \vec{v_1} . \vec{v_2} = 0 $

$\large (u_1 \hat{i} – g t\hat{j}) . (- u_2 \hat{i} – g t\hat{j}) = 0 $

$\large u_1 u_2 = g^2 t^2 $

$\large t = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large \sqrt{\frac{2 h}{g}} = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large h = \frac{u_1 u_2}{g}$

Q: A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. if the gun can impart a velocity of 500 m/sec to the bullet, at what height above the bird must he aim gun in order to hit it ?

Sol. x = vt or 100 = 500 × t ; t = 0.2 sec.

Now h = 0 + (1/2) × 10 × (0.2)²

= 0.20 m = 20 cm

Q: A golfer standing on the ground hits a ball with a velocity 52 m/s at an angle θ Above the horizontal if tan⁡θ =  5/12 find the time for which the ball is at least 15 m above the ground ? (g = 10 m/s² )

Sol:  $\large u_y = u sin\theta$

$\large v_y = \sqrt{u_y^2 – 2 g y }$

$\large v_y = \sqrt{(u sin\theta)^2 – 2 g y }$

$\large v_y = \sqrt{(52 \times \frac{5}{13})^2 – 2 \times 10 \times 15 }$

v_y = 10 m/s

$\large \Delta t = \frac{2 v_y}{10} = \frac{2 \times 10}{10} $

= 2 sec

Q: In the absence of wind the range and maximum height of a projectile were R and H. If wind imparts of horizontal acceleration a = g/4 to the projectile then find the maximum range and maximum height.

Sol. H’ = H (since , u sin⁡θ remain same )

T’ = T

$\large R’ = u_x T + \frac{1}{2} a T^2$

$\large R’ = R + \frac{1}{2} (\frac{g}{4}) T^2$

$\large R’ = R + \frac{1}{2} g T^2$

R’  = R + H (since H’ =H)