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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 2.3o3logm_o/m

Where m_o/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 = F_av x t [ t is the time of impact]

F_av = Impulse/t

F_av = 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 F_av = 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 M_c = 5.71 x 10²² kg

Initial velocity of comet  u_c = 1.41 x 10⁴ m/s

Mass of earth M_e = 5.98 x 10²⁴ kg

Initial velocity of earth  u_e = 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 M_c u_c =( M_e + M_c) v

where v is the final velocity of both earth and comet.

hence

v = M_c u_c / ( M_e + M_c) = 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.)