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Ne4ueva [31]
3 years ago
12

in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

ve together. What is the total momentum after the collision? 55 kg*m/s 275 kg * m/s -55kg * m/s -275 kg m/2
Physics
1 answer:
guajiro [1.7K]3 years ago
4 0

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

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Answer:

6\:\mathrm{m/s}

Explanation:

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Plugging in given values, we get:

S=\frac{30}{5}=\fbox{$6\:\mathrm{m/s}$}.

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You are in a car moving forward at 12 m/s. You throw a ball in the direction the car is moving. From your point of view, what do
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A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
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Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

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Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

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d₁ = 0 + 1/2 × at₁²

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d₁ = 109.5 m

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d_{c} = 22.352 × 7

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t₂ = Δd / ( V_{f} - V_{c} )

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t₂ = 46.96 / 8.938

t₂ = 5.25 sec

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d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

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a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

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