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CaHeK987 [17]
2 years ago
11

A 60-kg skate boarder is standing on his stationary board. A bigger, out of control 75-kg skate boarder crashes into him with a

velocity of 4 m/s. Their boards get stuck together in this inelastic collision. How fast are the moving after the collision?
Physics
1 answer:
andrezito [222]2 years ago
3 0

Answer:

2.22m/s

Explanation:

Given parameters:

M1  = 60kg

M2  = 75kg

V1  = 0m/s

V2  = 4m/s

Unknown;

Velocity after collision  = ?

Solution:

To solve this problem, we must understand that the momentum before and after collision of the bodies must be the same;

     M1 V1  + M2 V2  = v(M1 + M2)

So;

       60 x 0  +  75 x 4  = v (60 +75)

                 300  = 135v

                   v = 2.22m/s

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Read 2 more answers
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

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Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

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0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

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