Question

A ball of mass 5 kg is moving with velocity 4ms-1 in and comes to rest after travelling a distance of 4m. The force exerted on an object is ans fast plsssssss

Answer 1

Answer:

10N opposing the motion

Explanation:

v=0

u=4m/s

s=4m

Using Newton's equation of motion,

v^2 - u^2 = 2as

a = (v^2 - u^2)/2s

   = (0^2 - 4^2)/2*4

   =-2m/s^2

F=ma = (-2)*5 = -10N

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Answer 2

Let their velocities after the collision be v  1  and v  2

As we know for elastic collision is 

Relative velocity of approach = relative velocity of separation

10−4=v 2 −v 1

⇒6=v 2 −v 1

​⇒v  1  =v  2  −6

Applying conservation of momentum we will get,

10×10+5×4=10v  1 +5v  2

120=10v 1+5v 2

120=10(v  2−6)+5v  2 =15v 2 −60

15v  2  =180⇒v2  =12cm/sec

v  1  =6cm/sec

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