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A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is
Impulse experienced by the body= change in momentum = MV – (–MV) = 2MV.
A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to
\(F=\frac{m(v-u)}{t}=\frac{0.15(0-20)}{0.1}=30\: N\)
A body of mass 1.0 kg is falling with an acceleration of 10 m/sec2. Its apparent weight will be (g = 10 m/sec2)
Apparent weight when mass is falling down is given by W'=m(g-a)
∴W'=1×(10-10)=0
In an explosion, a body breaks up into two pieces of unequal masses. In this
If m1, m2 are masses and u1, u2 are velocity then by conservation of momentum m1u1 + m2u2= 0 or
|m1u1| = |m2u2|
The minimum velocity (in ms-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
The condition to avoid skidding, v =\(\sqrt{urg}=\sqrt{0.6\times 150\times 10}=30 m/s\)
A rocket of mass 5000 kg is to be projected vertically upward.The gases are exhausted vertically downwards with velocity 1000 ms-2 with respect to the rocket. What is the minimum rate of burning the fuel so as to just lift the rocket upwards against gravitational attraction?
\(\frac{dm}{dt}=\frac{mg}{v_{r}}=\frac{5000\times 9.8}{1000}=49kgs^{-1}\)