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Whitepunk [10]
3 years ago
14

65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!

Physics
1 answer:
otez555 [7]3 years ago
3 0
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



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1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

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