Question

A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change in the momentum of the ball during the rebound?
A) 4.4 kg. m/s
B) 72 kg m/s
C) 1.2 kg.m/s
D) 1.8 kg m/s
E) 5.4 kg.m/s

Answer 1

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball $v_1=25m/sec$

And velocity after rebound $v_2=-19m/sec$

We have to find the change in momentum

So change in momentum is equal to $=m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec$ ( here negative sign shows only direction )

So option (A) will be correct answer

Answer 2

This question involves the concepts of momentum, mass, and velocity.

The magnitude of the change in momentum of the ball during the rebound is "A. 4.4 kg.m/s".

The change in momentum can be found using the following formula:

$\Delta P = P_i-P_f\\\Delta P=m(v_f-v_i)$

where,

ΔP = change in momentum = ?

m = mass of ball = 0.1 kg

vi = initial velocity = 25 m/s

vf = final velocity = - 19 m/s (negative sign due to opposite direction)

Therefore,

ΔP = (0.1 kg)(-19 m/s - 25 m/s)

ΔP = - 4.4 kg.m/s (negative sign shows the direction)

Learn more about momentum here:

https://brainly.com/question/14878836?referrer=searchResults