Question

Two friends of equal mass 45kg each are standing on smooth surface. A throws ball of mass 5 kg
towards B along the surface and B catches it. Choose correct options (Ball moves at 10 m/s w.rit.
surface.)
(A) Speed of A after the throws is 1 m/s (B) Speed of A after he throws is 1.11 m/s
(C) Speed of B after be catches id 1 m/s (D) Speed of B after he catches is 1.11 m/s

Answer 1

Answer:

(B) Speed of A after he throws is 1.11 m/s

(C) Speed of B after be catches id 1 m/s

Explanation:

Here we can use momentum conservation as there is no external force on the system of ball and the person

So we will have

$P_i = P_f$

here we know that

$P_i$ = initial momentum of the ball + person A

$P_f$ = final momentum of ball + person A

so we have

$45(0) + 5(0) = 45(v_a) + 5(10)$

$v_a = - 1.11 m/s$

so speed of A is 1.11 m/s in opposite to the velocity of ball

now we can again use momentum conservation for ball and person B

$5(10) + 45(0) = (5 + 45) v$

$50 = 50 v$

$v = 1m/s$

#Learn

Topic : Momentum conservation

https://brainly.in/question/3576502