A soccer ball of mass 250 text{ g} is moving | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: Mass $m = 250 	ext{ g} = 0.25 	ext{ kg} = \frac{1}{4} 	ext{ kg}$.Initial velocity $\vec{u} = -22 \hat{i} 	ext{ ms}^{-1}$.Final velocity

Vedclass · Accepted Answer

$1166$

Vedclass · Answer

(C) Given: Mass $m = 250 	ext{ g} = 0.25 	ext{ kg} = \frac{1}{4} 	ext{ kg}$.Initial velocity $\vec{u} = -22 \hat{i} 	ext{ ms}^{-1}$.Final velocity $\vec{v} = 30 \cos 53^{\circ} \hat{i} + 30 \sin 53^{\circ} \hat{j} = 30(\frac{3}{5}) \hat{i} + 30(\frac{4}{5}) \hat{j} = 18 \hat{i} + 24 \hat{j} 	ext{ ms}^{-1}$.Change in momentum $\Delta \vec{P} = m(\vec{v} - \vec{u}) = \frac{1}{4} [(18 \hat{i} + 24 \hat{j}) - (-22 \hat{i})] = \frac{1}{4} [40 \hat{i} + 24 \hat{j}] = 10 \hat{i} + 6 \hat{j} 	ext{ Ns}$.Magnitude of change in momentum $|\Delta \vec{P}| = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} 	ext{ Ns}$.Average force $\vec{F}_{avg} = \frac{\Delta \vec{P}}{\Delta t} = \frac{\sqrt{136}}{0.01} = 100 \sqrt{136} \approx 100 	imes 11.66 = 1166 	ext{ N}$.

Similar Questions

$A$ machine gun fires a bullet of mass $65 \, g$ with a velocity of $1300 \, m/s$. The man holding it can exert a maximum force of $169 \, N$ on the gun. The number of bullets he can fire per second will be ............

$A$ cricket ball of mass $0.25 \,kg$ with speed $10 \,m/s$ collides with a bat and returns with the same speed within $0.01 \,s$. The force exerted on the bat is: (in $\,N$)

When the speed of a moving body is doubled:

The momentum $p$ of an object varies with time $(t)$ as shown in the figure. The corresponding force $(F)$-time $(t)$ graph is

Vedclass Test Series

Exam Paper Generator

Online Exam Module