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Fudgin [204]
3 years ago
7

A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s.

Take right to be the positive direction. What is the total momentum of the two balls before they collide? What is the total momentum of the two balls after they collide?
Physics
1 answer:
Savatey [412]3 years ago
5 0

Explanation:

Momentum before collision:

(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s

No external forces act on the balls, so momentum is conserved.  Therefore, momentum after collision is also -3 kg m/s.

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A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.60 kg is attached to the sprin
trasher [3.6K]

(a) 1.08 J

The elastic potential energy stored in the block at any position x is given by

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement relative to the equilibrium position

Here we have

k = 860 N/m

x = 5.00 cm = 0.05 m is the position of the block

Substituting, we find

U=\frac{1}{2}(860 N/m)(0.05 m)^2=1.08 J

(b) 1.16 m/s

The total mechanical energy of the spring-mass system is equal to the potential energy found at point (a), because there the system was at its maximum displacement, where the kinetic energy (because the speed is zero).

At the equilibrium position, the mechanical energy is sum of kinetic and potential energy

E = K + U

However, at equilibrium position x = 0, so U = 0. Therefore, the kinetic energy is equal to the total energy found at point (a)

E=K= \frac{1}{2}mv^2 = 1.08 J

where

m = 1.60 kg is the mass of the block

v is the speed

Solving for v, we find

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.08 J)}{1.60 kg}}=1.16 m/s

(c) 1.00 m/s

When the block is at position x = 2.50 cm, the mechanical energy is sum of kinetic and potential energy:

E=K+U=\frac{1}{2}mv^2 + \frac{1}{2}kx^2

where

E = 1.08 J is the total mechanical energy

m = 1.60 kg is the mass

v is the speed

k = 860 N/m

x = 2.50 cm = 0.025 m is the displacement

Solving for v, we find

v = \sqrt{\frac{2E - kx^2}{m}}=\sqrt{\frac{2(1.08 J)-(860 N/m)(0.025 m)^2}{1.60 kg}}=1.00 m/s

8 0
3 years ago
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Explanation:

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My answer is 6 watts because 30J/5s is 6
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Which of the below statements are true?
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Answer:

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