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sleet_krkn [62]
3 years ago
8

A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra

in is moving in the positive x-direction with a velocity of Vi = 2.45 m/s and the toy car is also moving in the positive x-direction with a velocity of vi = 4.75 m/s. Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 2.15 m/s.
(a) Determine Vf, the final velocity of the toy train.
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

  v_{f} = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

    p₀ = m v₁₀ + M v₂₀

After the inelastic shock

    p_{f}= m v_{1f} + M v_{2f}

    p₀ =  p_{f}

    m v₀ + M v₂₀ = m v_{1f}  + M v_{2f}

We cleared the end of the train

     M v_{2f} = m (v₁₀ - v1f) + M v₂₀

Let's calculate

     3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

     v_{2f}  = (-3.9 + 8.82) /3.60

      v_{2f}  = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

      m v₁₀ + M v₂₀ = (m + M) v_{f}

      v_{f}  = (m v₁₀ + M v₂₀) / (m + M)

      v_{f}  = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

      v_{f} = 3,126 m / s

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Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

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Given:

v(final velocity)= 7.5 m/s

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s=ut+1/2(at^2)

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u is the initial velocity measured in m/sec

a is the acceleration measured in m/s^2.

t is the time taken to cover this distance.

Substituting the given values in the above formula we get

19= 4.5u+1/2(a x 4.5 x 4.5)

20.25 a + 9 u = 38


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v= u + at

Substituting the given values in the above formula we get

7.5= u + 4.5a


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The Canadian bobsled team hit the brakes on the sled they are traveling East in so that it decelerates at a rate of 0.43
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Answer:

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