A 4 kg bowling ball moving at 1.4 m/s east impacts a 400 g pin that is stationary. After the impact, the ball is moving at 0.5 m/s east. Assuming the collision was elastic, what speed does the pin move at after being struck?



Answer :

The speed of the pin after the elastic collision is 9 m/s east.

Final speed of the pin

The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;

m1u1 + mu2 = m1v1 + m2v2

where;

  • m is the mass of the objects
  • u is the initial speed of the objects
  • v is the final speed of the objects

4(1.4) + 0.4(0) = 4(0.5) + 0.4v2

5.6 = 2 + 0.4v2

5.6 - 2 = 0.4v2

3.6 = 0.4v2

v2 = 3.6/0.4

v2 = 9 m/s

Thus, The speed of the pin after the elastic collision is 9 m/s east.

Learn more about linear momentum here: https://brainly.com/question/7538238

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