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schepotkina [342]
3 years ago
14

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

ocity of 2.60 m/s, grabs and holds onto him so that they move off together with a velocity of 1.30 m/s. If the mass of the tackler is 122 kg, determine the mass of the receiver. Assume momentum is conserved.
Physics
1 answer:
amm18123 years ago
5 0

Answer:

<h2>122kg</h2>

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

<u>Before collision;</u>

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

<u>After collision</u>

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

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A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o
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a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

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         W = K_f - K₀

          W = ½ m v_f² - ½ m v₀²

         W = ½ 0.325 (5.5² - 8.5²)

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b) find us the coefficient of friction

Let's use Newton's second law

            fr = μ N

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            fr = μ W

work is defined by

             W = F d

the distance traveled in a revolution is

             d₀ = 2π r

             W = μ mg d₀ = -6.825

            μ = \frac{ -6.825}{d_o \ mg}

               

The total work as the object stops the final velocity is zero v_f = 0

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           μ mg d = -11.74

           

we subtitle the friction coefficient value

           ( \frac{-6.8525 }{d_o mg}) m g d = -11.74

               6.825  \frac{d}{d_o} = 11.74

               d = 11.74/6.825  d₀

               d = 1.7201  2π 0.400

               d = 4.32 m

this is the total distance traveled, the distance and the angle are related

              θ = d / r

              θ = 4.32 / 0.40

              θ = 10.808 rad

we reduce to revolutions

              θ = 10.808 rad (1rev / 2π rad)

              θ = 1.72 revolution

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