Elasticity Laws of Motion One Dimension Motion

Question 1

A120 lb person is being lifted aloft by a rocket which accelerates uniformly to an upward speed of 300 ft/sec in 2 sec. What is the upward acceleration in multiples of g?

Hint 1

Acceleration of the rocket can be calculated from first equation of motion v = u + at

Answer 1

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

Question 2

A120 lb person is being lifted aloft by a rocket which accelerates uniformly to an upward speed of 300 ft/sec in 2 sec. 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.

Hint 2

mg - Actual weight of the man

ma = N - mg

N = ma + mg

Answer 2

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.

Question 3

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.

Hint 3

Find the force applied and the acceleration of the toy and the string. From that find the force applied on the toy

Answer 3

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

Question 4

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.

Hint 4

Find the final mass of the rocket when the gasses have exhausted. Then use the ratio of the initial mass of the rocket and the mass at instant of time to find the velocity

Answer 4

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

Question 5

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.

Hint 5

When the acrobat bounces back the impulse acting on the acrobat = Change in momentum = m [v-(-v)] = 2mv

Answer 5

Physics Practice and Skills Development Worksheets

Elasticity Laws of Motion One Dimension Motion

Question 1

Acceleration of the rocket can be calculated from first equation of motion v = u + at

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

Question 2

mg - Actual weight of the man

ma = N - mg

N = ma + mg

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.

Question 3

Find the force applied and the acceleration of the toy and the string. From that find the force applied on the toy

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

Question 4

Find the final mass of the rocket when the gasses have exhausted. Then use the ratio of the initial mass of the rocket and the mass at instant of time to find the velocity

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

Question 5

When the acrobat bounces back the impulse acting on the acrobat = Change in momentum = m [v-(-v)] = 2mv

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

a = 150 ft/sec²

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

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

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

v² = u² + 2gh

Impulse = F_av x t [ t is the time of impact]

F_av = Impulse/t

F_av = 1292.93/t