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A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
to the west. The launching rod strikes the pinball and causes it to move in the
opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the
pinball by the launcher?
A 0.75 kg•m/s to the east
B 1.2 kg•m/s to the east
C 1.8 kg•m/s to the east
D 3.0 kg•m/s to the east

Respuesta :

i believe the answer is option C 1.8 kg•m/s to the east

The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east

Impulse:

Let us consider east as positive direction and west as negative direction.

Then, from the question, we get that the initial velocity of the pinball is

2m/s towards the west

or u = - 2 m/s

and the mass of the pinball is m = 150g = 0.15 kg

So, the initial momentum of the pinball is:

[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]

Now, the final velocity of the pinball after being struck by the rod is 10  m/s towards the east,

or v = 10 m/s

So, the final momentum of the pinball is:

[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]

Impulse is defined as the change in momentum, that is,

[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]

The impulse is 1.8 kgm/s towards the east.

Learn more about impulse:

https://brainly.com/question/14848283?referrer=searchResults

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