A ball of mass 0.2 kg is thrown from a heigh | vedclass

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Question

(C) The force acting on the ball is gravity, which acts only in the vertical $(y)$ direction. Therefore, there is no change in momentum in the horizon

Vedclass · Accepted Answer

$\frac{1+\sqrt{5}}{\sqrt{5}}$

Vedclass · Answer

(C) The force acting on the ball is gravity, which acts only in the vertical $(y)$ direction. Therefore, there is no change in momentum in the horizontal $(x)$ direction.Initial vertical velocity: $u_y = u \sin 45^{\circ} = \sqrt{10} \cdot \frac{1}{\sqrt{2}} = \sqrt{5} \ m/s$.Initial vertical momentum: $p_{yi} = m u_y = 0.2 \cdot \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}} \ kg \cdot m/s$ (upwards).Using the equation of motion $v_y^2 = u_y^2 + 2gh$ (taking downward as positive for final velocity calculation, or using energy conservation):$v_y^2 = (\sqrt{5})^2 + 2(10)(1) = 5 + 20 = 25$.So, final vertical velocity $v_y = 5 \ m/s$ (downwards).Final vertical momentum: $p_{yf} = m v_y = 0.2 \cdot 5 = 1 \ kg \cdot m/s$ (downwards).Taking upward direction as positive, $p_{yi} = \frac{1}{\sqrt{5}}$ and $p_{yf} = -1$.The change in momentum is $\Delta p = p_{yf} - p_{yi} = -1 - \frac{1}{\sqrt{5}} = -(1 + \frac{1}{\sqrt{5}})$.The modulus of the momentum increment is $|\Delta p| = 1 + \frac{1}{\sqrt{5}} = \frac{\sqrt{5}+1}{\sqrt{5}}$.

List-$I$ (Kinematic Variable)	List-$II$ (Value)
$A$. Velocity of the body after $1 \ s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \ s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \ s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \ s$ (in $ms^{-1}$)	$IV$. $22.4$

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