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Margaret [11]
3 years ago
9

A 1400 kg car at rest at a stop sign is rear ended by a 1860 kg truck traveling at a speed of 19.1 m/s. After the collision, the

two vehicles are locked together. Alice is moving past the collision site in a frame S' that is traveling at a constant velocity of 6.11 m/s in the direction of travel of the incident truck.
(a) Determine the momentum of the vehicles before the collision in Alice's frame. kg · m/s

(b) Determine the momentum of the vehicles after the collision in Alice's frame. kg · m/s

(c) Does Alice conclude that momentum is conserved during the collision? Yes No cannot be determined
Physics
1 answer:
Anna71 [15]3 years ago
8 0

Answer:

Explanation:

Applying conservation of momentum law to system of car and truck

their common velocity = their total momentum before collision / total mass

= (0 + 1860 x 19.1 ) / (1400 + 1860)

= 10.8975  m/s

Velocity of car with respect to ground before the collision u₁ = 0

velocity of the car in the frame of Alice                 u₁ = 0 - 6.11 = - 6.11 m /s

similarly velocity of truck in the frame of Alice = 19.1 - 6.11 = 12.99 m /s

total momentum of system of car and truck in alice's frame

= m₁ u₁ + m₂u₂  , m₁ , m₂ are masses of car and truck and u₁ , u₂ are their velocities in Alice's frame

= 1400 x - 6.11 + 1860 x 12.99

= - 8554 + 24161.4

= 15607.4 kg m/s

velocity of car and truck  after collision in Alice's frame

= 10.8975 - 6.11

= 4.7875

total momentum after collision in Alice's frame

= 4.7875 x ( 1400 + 1860 )

= 15607.25

Total momentum before collision in Alice's frame

= Total momentum in Alice's frame after collision

So law of conservation of momentum is obeyed in Alice's frame also.

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Inserting the values

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two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
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Answer:

t=750s

Explanation:

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\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
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(a) 1200 rad/s

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\alpha = \frac{\omega_f - \omega_i}{t}

where we have

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Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

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If we re-arrange it for t, we get:

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