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 pla ans fast mak barinliest

Answer 1

Answer:

Step-by-step explanation:

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

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As we know for elastic collision.  

Relative velocity of approach = relative velocity of separation

10−4=v 2 −v 1

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⇒6=v 2 −v 1

​⇒v  1  =v  2  −6

Applying conservation of momentum,

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