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In order to complete a tough "split" in bowling, a bowling ball of mass 6.5 kg is rolled with a velocity of 12 m/s. the ball hits a 1.3 kg bowling pin, sending it off with a speed of 3.5 m/s in the same direction of the original direction of the bowling ball. what is the bowling ball’s velocity after hitting the pin?

User Whiterose
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The question involves finding the post-collision velocity of a bowling ball after hitting a pin, using conservation of linear momentum. The mass and initial velocity of both the ball and the pin are given, and by applying the principle, the final velocity of the bowling ball can be calculated.

The student is asking about the velocity of a bowling ball after it has hit a bowling pin, a question that can be answered using the principle of conservation of momentum in physics. To find the post-collision velocity of the bowling ball, we apply the conservation of linear momentum which states that the total momentum of an isolated system remains constant if there is no external force acting on it.

In mathematical terms, for a collision between two objects, this principle can be expressed as m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses, v1 and v2 are the initial velocities, and v1' and v2' are the final velocities of the two objects, respectively.

For the given problem, we have a bowling ball of mass 6.5 kg with an initial velocity of 12 m/s hitting a bowling pin of mass 1.3 kg which ends up moving at 3.5 m/s in the same direction. By substituting the known values into the conservation of momentum equation, we can calculate the final velocity of the bowling ball after the collision.

In conclusion, by utilizing the conservation of momentum, we apply the equation: (6.5 kg * 12 m/s) + (1.3 kg * 0 m/s) = (6.5 kg * v') + (1.3 kg * 3.5 m/s), and solving for v'. Once calculated, this value gives us the final answer for the velocity of the bowling ball post-collision.

User Krsnaa
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