Jake is bowling with an 8 kg bowling ball roll the ball at 7 ms and it hits 1 stationary pain with a mass of 2kg the pain goes flying forward at 12 ms how fast is the ball now traveling

Solution

In this question we have given

mass of bowling ball=8kg

Velocity of bowling ball= 7m/s

mass of pain=2kg

velocity of pain=12m/s

In this case, law of conservation of momentum will be applied. According to law of conservation of momentum,

initial momentum of system= final momentum of the system

Total initial momentum of the system =mass of bowling ball* Velocity of bowling ball

=8 kg x 7 m/s

= 56 kg m/s.

momentum of the pain after getting hit is =mass of pain* velocity of pain

=2kg*12m/s

=24 kg m/s.

So the momentum of the bowling ball after hitting the pain= 56 kg m/s - 24 kg m/s

= 32 kg m/s

velocity of the bowling ball after hitting the pain = momentum of the bowling ball after hitting the pain/mass of bowling ball

=32 kg m/s / 8 kg

= 4 m/s

Therfore, velocity of the bowling ball after hitting the pain is 4m/s

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.