Question 13: Work, Energy and Power & Circular Motion

Question

An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies at points $B$ and $C$ to that at point $A$ (see figure), in the absence of air friction is :

Accepted Answer

Let the mass of the object be m and the initial speed be u .
The angle of projection is $	heta=60^{\circ}$.
At point A
The kinetic energy at point $A$ is given as $K$.
$$ \mathrm{K}_{\mathrm{A}}=\frac{1}{2} \mathrm{mu}^2=\mathrm{K} $$
At Point B (Maximum Height)
At the highest point of a projectile's path, the vertical component of velocity becomes zero $\left(v_y=0ight)$. The object only possesses the horizontal component of velocity ( $\mathrm{v}_{\mathrm{x}}$ ).
The horizontal velocity remains constant throughout the motion (assuming no air resistance):
$$ \mathrm{v}_{\mathrm{B}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos 	heta $$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{B}}=\mathrm{u} \cos 60^{\circ}=\mathrm{u}\left(\frac{1}{2}ight)=\frac{\mathrm{u}}{2} $$
So, the kinetic energy at B is,
$$ \mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{B}}ight)^2 $$
$$\Rightarrow $$ $\mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}ight)^2$
$$\Rightarrow $$ $\mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}^2}{4}ight)$
$$\Rightarrow $$ $\mathrm{K}_{\mathrm{B}}=\frac{1}{4}\left(\frac{1}{2} \mathrm{mu}^2ight)=\frac{1}{4} \mathrm{~K}_{\mathrm{A}}=\frac{\mathrm{K}}{4}$
$$ \Rightarrow \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{K}}{4} $$
At Point C
Point C is at the same horizontal level as point A . By the law of conservation of energy, the speed of the projectile when it hits the ground at the same level is equal to the speed of projection.
$$ \mathrm{v}_{\mathrm{C}}=\mathrm{u} $$
Therefore, the kinetic energy at C is equal to the kinetic energy at A :
$$ \mathrm{K}_{\mathrm{C}}=\mathrm{K}_{\mathrm{A}}=\mathrm{K} $$
The difference in kinetic energies at B and C :
$$ \Delta K_{B C}=\left|K_C-K_Bight|=\left|K-\frac{K}{4}ight|=\frac{3 K}{4} $$
So, the ratio of the difference in kinetic energies at points B and C to that at point A is
Ratio $=\frac{\Delta K_{B C}}{K_A}=\frac{(3 / 4) K}{K}=\frac{3}{4}$
Hence, the correct option is (A).

Answer

$3: 4$

Answer

$1: 4$

Answer

$2: 3$

Answer

$1: 2$

Work, Energy and Power & Circular Motion

An object is projected with kinetic energy K from a point A at an angle 60^{\circ} with the horizontal. The ratio of the difference in kinetic energies at points B and C to that at point A (see figure), in the absence of air friction is :

Choose the correct answer: