Question

A force in the negative direction of an x axis is applied for 29 ms to a 0.25 kg ball initially moving at 13 m/s in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude 48.9 N s. (a) What is the ball's velocity (including sign for direction) just after the force is applied

Answer 1

Answer:

182.6m/s

Explanation:

Given:

collision time 't'= 27ms

mass of ball= 0.25kg

initial velocity 'vi'= 13m/s

Impulse magnitude'J' = 48.9Ns

the average force on the ball can be determined by formula during the collision i.e

Favg = J/ Δt

Favg= = -(Δm/Δt ) x ΔV

where,

J= impulse

Δt= collision time

ΔV= average velocity

Δm= average mass =m

therefore,

Vf- Vi = -J/m

Vf= -J/m + Vi

Vf= -(48.9/ 0.25) + 13

Vf= -182.6 m/s

|Vf| = 182.6m/s

therefore, the  the ball's velocity just after the force is applied is  182.6m/s