| taken from google.com Q. A120 lb person is being lifted aloft by a rocket which accelerates uniformly to an upward speed of 300 ft/sec in 2 sec. a) What is the upward acceleration in multiples of g? Mass of the man = 120lb Final speed of the rocket = 300 ft/sec Time = 2 sec Acceleration of the rocket can be calculated from first equation of motion v = u + at 300 = 0 + a x 2 a = 150 ft/sec2 = 4.7g b) What is the apparent weight of the person during acceleration? Free body diagram: N - Normal reaction which is equal to the apparent weight of the man. mg - Actual weight of the man ma = N - mg N = ma + mg = m (a + g) = m (4.7g + g) = 5.7mg = 5.7 x 120 g = 684g = 684 lbf. | |
Q. A short 10 gram string is used to pull a 50 gram toy across a frictionless horizontal surface. If a 3.0 x 10 Newton force is applied horizontally to the free end. Find the force of the string on the toy, at the other end.
Mass of the string = 10 grams = 10-2 kg
Mass of the toy = 50 gm = 5 x 10-2 kg
Mass of toy + string = 10-2 + 5 x 10-2 = 6 x 10-2 kg
Force applied F = 3.0 x 10 N
Acceleration of toy + string = 30/6x10-2 = 5 x 102 = 500 m/s2
Thus force on the toy = 5 x 10-2 x 500 = 25 N
Q. A rocket exhausts fuel with a velocity of 1500 m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80% of the total mass. Find its speed when all the fuel has been exhausted.
Exhaust speed of the gases = 1500 m/s = u
Initial mass of rocket + fuel = M
Mass of fuel = 80% of M
Final mass of rocket when the gases have been exhausted = 20%M = 0.2M
Velocity of the rocket at any time = v
v = u x 2.3o3logmo/m
Where mo/m is the ratio of the initial mass of the rocket to its mass at the instant of time
v = 1500 x 2.303 log M / 0,2M
= 2414.6 m/s
Q A 640 N acrobat falls 5.0 m from rest into a net. The net tosses him back up with the same speed he had just before he hit the net. Find the average force exerted on him by the net during this collision.
Weight of the acrobat = 640 N
Initial velocity of the acrobat = o
Height from which the acrobat falls = 5m
Speed of the acrobat when he hits the net = v
v2 = u2 + 2gh
= 0 + 2 x 9.8 x 5 = 98
v = v98 = 9.899 m/s
He bounces back with the same speed
therefore the velocity with which he bounces back = -9.899 m/s
Impulse acting on the acrobat = Change in momentum = m [v-(-v)] = 2mv
Mass of the acrobat = m = 640/9.8 = 65.306 kg
therefore impulse = 2 x 65.306 x 9.899 = 1292.93 Ns
Impulse = Fav x t [ t is the time of impact]
Fav = Impulse/t
Fav = 1292.93/t
Q. An automatic rifle fires 100 gram bullets at a speed of 571 m/s at a rate of 16 bullets per second. What is the average recoil force against the gun's mount?
Mass of bullet m = 100 grams = 0.1 kg
Velocity of bullet v = 571 m/s
Rate = 16 bullets / s
therefore Mass / s = 16 ´ 0.1 kg/s = 1.6 kg/s
Average force Fav = 1.6 ´ 571 = Dp / Dt = Dmv / Dt = v. Dm / Dt = 91.36 N
where Dp = small change in force and Dt = small change in time
The average recoil against the gun's mount = 91.36 N
Q. Suppose a comet of mass 5.71e+22kg moving at 1.41e+4m/s collides with the earth, initially at rest, and sticks to the earth. The mass of the earth is 5.98e+24 kg. Find the final speed of the earth.
A.
Mass of comet Mc = 5.71 x 1022 kg
Initial velocity of comet uc = 1.41 x 104 m/s
Mass of earth Me = 5.98 x 1024 kg
Initial velocity of earth ue = 0 m/s
After hitting the earth the comet sticks to it. The collision is in-elastic. This implies that linear momentum is conserved and kinetic energy is not conserved.
therefore Mc uc =( Me + Mc) v
where v is the final velocity of both earth and comet.
hence
v = Mc uc / ( Me + Mc) = 5.71 x 1022 x 1.41 x 104 / (5.98 x 1024 + 5.71 x 1022)
= 1.33 x 102 m/s
Q. A solid 1.56 kg sphere is rolling along a level surface, without slipping at 1.41 m/s. What is its kinetic energy?
A.
Moment of inertia of a solid sphere = 2/5 mr2 = I
Kinetic energy = kinetic energy of translation + kinetic energy of rotation
=½ mv2 + ½ I?2 = ½ mv2 + ½(2/5 mr2)( v2/ r2) as v = ?r
=½ mv2 + 1/5 mv2= 7/10 mv2 =7/10 x 1.56 x (1.41)2
= 2.17 J
Q Does a negative acceleration necessarily imply that a body is slowing down? What (or who) determines the sign if the acceleration in a particular problem?
A certain horse says, ?There?s no use trying to pull that cart. No matter how hard I pull on the cart, it will always pull back on me with exactly the same force. So I can never make it go.? Explain what is wrong with the horse?s reasoning. (Free-body diagrams, showing all the forces involved, will help.)
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