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Olegator [25]
3 years ago
9

1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of

10 m/s due north. The two vehicles stick together after the collision. a. What is the momentum of each vehicle prior to the collision? b. What are the size and direction of the total momentum of the two vehicles after they collide?
Physics
1 answer:
velikii [3]3 years ago
4 0

(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

Momentum of truck = Mass * Speed = 4000 * 20 =80000  kgm/s

Similarly, the momentum of car will be

Momentum of car = 1350 * 10 = 13500 kgm/s

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.

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6 0
3 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

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d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

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\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

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\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

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8 0
3 years ago
A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra
mr_godi [17]
<h2>Answer:</h2>

The rate of deceleration is -0.14m/s^{2}

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

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v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v =   3 minutes

<em>Convert the time t = 3 minutes to seconds;</em>

=> 3 minutes = 3 x 60 seconds = 180seconds.

<em>Substitute the values of v, u, t into the equation above. We have;</em>

v =  u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

<em>Make a the subject of the formula;</em>

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14m/s^{2}

The negative value of a shows that the boat is decelerating.

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Therefore; the minimum applied force will be 97.6 Newtons.
4 0
3 years ago
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