Question

 smooth sphere of mass 1 kg is moving with horizontal speed of 3 m/s strikes perpendicularly with a vertical wall and bounces off the wall with horizontal speed of 2 m/s. What is the coefficient of restitution between sphere and wall?​

Answer 1

Answer:

Your answer is 2/3

Explanation:

coefficient of restituion is

e = separation speed / approach speed

e = 2-0 / 3-0

e = 2/3

       

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Answer 2

Answer:

The coefficient of restitution between sphere and ball is 0.66.

Explanation:

Here,

The velocity of approach between the wall and ball is denoted by $V_{app}$.

The velocity of separation between the ball and wall is denoted by $V_{sep}$.

The coefficient of restitution between sphere and wall is denoted by e.

The velocity of the ball is denoted by $V_{b}$.

The velocity of the wall is denoted by $V_{w}$.

Now,

$V_{app} =V_{b}-V_{w}$

       = 2-0

       = 2m/sec

$V_{sep} =V_{b}-V_{w}$

      = 3-0

      = 3m/sec

Then,

By the equation,

  $e=\frac{V_{sep} }{V_{app} }$

  $e= \frac{2}{3}$

  e = 0.66

So, the coefficient of restitution between sphere and ball is equal to 0.66.