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4-1.Newton's Laws of Motion
hard
A ball is thrown up at an angle with the horizontal. Then, the total change of momentum by the instant it returns to the ground is
A
acceleration due to gravity $\times$ total time of flight
B
weight of the ball $\times$ half the time of flight
C
weight of the ball $\times$ total time of flight
D
weight of the ball $\times$ horizontal range
Solution
Change in momentum of the ball,
$\Delta \mathrm{p}=\mathrm{mv} \sin \theta-(-\mathrm{mv} \sin \theta)=2 \mathrm{mv} \sin \theta$
Where $\mathrm{v}$ is the velocity of prjection and $\theta$ is the angle of projection with the horizontal and $\mathrm{m}$ is the mass of the ball.
$\Delta p=2 m g v \times \frac{\sin \theta}{g}$
$=m g \times \frac{2 v \sin \theta}{g}$
$=$ weight of the ball $\times$ total time of flight.
Standard 11
Physics
