Question

Ball A is rolled in a straight line with a speed of 5m/s towards a bigger ball B lying 20m away. After collision with ball B, ball A retraces it's path and reaches its starting point with a speed of 4m/'s. What is the average velocity of ball A during the time interval 0 to 6s?

Answer 1

Answer: 0 m/s

Explanation:

SINCE THE BALLs STARTING AND ENDING POSITION IS SAME.

THEREFORE THE DISPLACEMENT =0.

AVERAGE VELOCITY = DISPLACEMENT/TIME

                           =0/TIME

                            =0 m/s

Answer 2

Answer:

Concept:

A vector quantity is average velocity. The change in stance or displacement (x) multiplied by the time intervals (t) in which the displacement happens is the average velocity. Depending on the sign of the displacement, the average velocity can be positive or negative. Meters every second (m/s or ms-1) is the SI measure for average velocity.

Given:

Ball A is thrown in a straight path at a speed of 5 metres per second towards a larger ball B, which is 20 metres away. Ball A retraces its journey after colliding with ball B and arrives at its starting location at a speed of 4m/s.

Find:

During the time span 0 to 6s, what is the mean speed of ball A?

Answer:

The average velocity can be stated as a relationship between the total time necessary for the journey and the total distance travelled.

$V(avg)=S(tot)/t(tot)$

It is necessary to compute the entire time in 6 seconds and the overall time

$t(AB)=\frac{S(AB)}{v(AB)}$

          $=\frac{20m}{5m/s}$

         $= 4 sec$

As a result, the ball will travel from B to some position C between B and A during the next 2 seconds:

$S(BC)= t(BC)* v(BC)$

           $= 2 sec* 4 m/s$

           $= 8$

The overall distance for the 6 seconds will be 28 metres.

$V(avg)=\frac{S(tot)}{t(tot)}$

             $=\frac{28m}{6s}$

            $= 4.67m/s$

∴ the average velocity of ball A during the time interval 4.67 m/s

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