A bullet is fired from a gun. The force on the bullet is given by F = 600 – 2 × 105 t Where,F is in new tons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?
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Solution
A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m/s. To givean initial upward acceleration of 20 m/s2,the amount of gas ejected per second to supply the needed thrust will be(Take g= 10 m/s2)
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Solution
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m.
What is the work done by the force F? (Note :sin 60° = 0.87;cos 60° = 0.50. Ignore friction and the weights of the pulleys)
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Solution
Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force or the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5 m =200 J.
A 50 kg ice skater, initially at rest, throws a 0.15 kg snowball with a speed of 35 m/s. What is the approximate recoil speed of the skater?
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Solution
Momentum is always conserved. Since the skater and snowball are initially at rest, the initial momentum is zero. Therefore, the final momentum after the toss must also be zero.
Blocks A and B of masses 15 kg and 10 kg, respectively, are connected by a light cable passing over a friction less pulley as shown below. Approximately what is the acceleration experienced by the system?
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Solution
Two external forces, FA and FB, act on the system and move in opposite direction. Let’s arbitrarily assume that the downward direction is positive and that FA provides downward motion while FB provides upward motion.
FA= (+15 kg) (9.8 m/s2) = 147 N
and FB= (–10 kg)(9.8 m/s2) = – 98 N
Ftotal= FA + FB = 147 N + (–98 N) = 49 N
The total mass that must be set in motion is 15 kg + 10 kg = 25 kg
Since ,Ftotal=mtotala,a=Ftotal/mtotal
= 49 N / 25 kg ≅ 2 m/s2
A block is kept on a inclined plane of inclination θ of length ι.The velocity of particle at the bottom of inclined is (the coefficient of friction is μ)
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Solution
A particle starts sliding down a friction less inclined plane.If Sn is the distance traveled by it from time t= n – 1 sec tot = n sec, the ratio Sn/Sn+1 is
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Solution
A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in
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Solution
At the highest point of the track, N + mg=mv'2⁄r
where r is the radius of curvature at that point and v' is the speed of the block at that point.
Now N=mv'2⁄r-mg
N will be maximum when r is minimum (v' is the same for all cases). Of the given tracks, (a) has the smallest radius of curvature at the highest point.
