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1 year ago

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A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 2

5.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball

1 answer:

Luda [366]1 year ago

4 0

Answer:

0.08m/s

Explanation:

Given data

M1= 69kg

v1= 2.61m/s

M2= 0.22kg

v2= 25.2m/s

Before snowball is thrown:

Total mass of skater + snowball = 69+ 0.22 = 69.22kg

Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s

After snowball is thrown:

Let's call the velocity of the skater V.

Total momentum = momentum of skater + momentum of snowball

=69.22V + (5.544)

= 69.22V + 5.544

So:

180.7 = 69.22V+5.544

180.7- 5.544= 69.22V

175.156= 69.22V

V= 175.156/69.22

V = 2.53m/s

The total momentum after catching the snowball is mV or:

(69.0 + 0.22) x V

So:

5.544= 69.22V

V= 5.544/69.22

V=0.08m/s

The velocity of the ice skater after throwing the snowball is 0.08m/s

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Answer:

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We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

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B = Magnetic field strength

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Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

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