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nikitadnepr [17]
3 years ago
11

WILL MARK AS BRAINLIEST!!!!!!

Physics
1 answer:
yaroslaw [1]3 years ago
4 0

M1V1 + M2V2 = M1V1' + M2V2'

where:

M1 is the mass of the large marble = 0.05 kg

V1 is the initial velocity of the large marble = 0.6 m/sec

M2 is the mass of the small marble = 0.03 kg

V2 is the initial velocity of the small marble = 0 m/sec (marble is at rest)

V1' is the final velocity of the large marble = -0.2 m/sec

V2' is the final velocity of the small marble that we want to calculate

Substitute with the givens in the above equation to get V2' as follows:

M1V1 + M2V2 = M1V1' + M2V2'

(0.05)(0.6) + (0.03)(0) = (0.05)(-0.2) + 0.03V2'

0.03 = -0.01 + 0.03V2'

0.03V2' = 0.03+0.01 = 0.04

V2' = 0.04/0.03

V2' = 1.334 m/sec

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Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

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(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

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*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

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     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

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