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adoni [48]
3 years ago
15

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ection at 15.0 cm/s. after the collision, the 10.0 g marble moves to the right at 22.1 cm/s. find the velocity of the 25.0 g marble after the collision.
Physics
1 answer:
ikadub [295]3 years ago
6 0
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
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This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag
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Answer:

Part a)

F = 135.7 N

Part b)

F = 62.5 N

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg + F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg + \mu(Fsin\theta)

F(cos\theta - \mu sin\theta) = mg

F = \frac{mg}{cos\theta - \mu sin\theta}

F = \frac{55}{cos50 - 0.310(sin50)}

F = 135.7 N

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg - F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg - \mu(Fsin\theta)

F(cos\theta + \mu sin\theta) = mg

F = \frac{mg}{cos\theta + \mu sin\theta}

F = \frac{55}{cos50 + 0.310(sin50)}

F = 62.5 N

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3 years ago
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