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Exam3ReviewSp08

# Exam3ReviewSp08 - Exam 3 Review Exam 3 Monday Apr 28 Chaps...

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Exam 3 Review Exam 3 Monday Apr. 28 Chaps. 9,10 & 11

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Linear Momentum n The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity: n p = m v n The terms momentum and linear momentum will be used interchangeably in the text
Newton and Momentum n Newton called the product m v the quantity of motion of the particle n Newton’s Second Law with constant mass (i.e. dm/dt = 0) can be used to relate the momentum of a particle to the resultant force acting on it This is a generalization of Newton’s 2nd Law, useful when the mass is not constant.

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Ballistic Cannon A projectile (M p ) is fired from a cannon cart (M cc ). The projectile lands in the receiver cart (M rc ). In this DEMO M cc = M p + M rc There are no horizontal external forces. Thus the total momentum remains constant.
Ballistic Cannon Initially the momentum of the system is zero. Since only internal forces are involved the total system momentum stays zero. p = 0 After the cannon fires and before the projectile lands in the receiver cart: M cc V cc + M p V p = 0 V cc = -(M p /M cc )V p When the projectile lands in the receiver cart the momentum is still zero and the cannon cart is still going to the left. 0 = M cc V cc + (M p +M rc )V rc Since M cc = M p + M rc , V cc = - V rc

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Impulse and Momentum n From Newton’s Second Law, F = d p / dt n Solving for d p gives d p = F dt n Integrating to find the change in momentum over some time interval n The integral is called the impulse, I , of the force F acting on an object over D t D p = p f - p i = F dt t i t f = J
Problem- Deformed Ball A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose that photographs show the maximum depth of the dent is 1 cm. Compute an estimate for the acceleration of the ball while it is in contact with the pavement and the time duration it is deformed.

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Solution- Deformed Ball To find the contact time we can use the Impulse equation using the average force. To find the average force we use kinematics to find the average acceleration, assuming constant a gives an approximate average a. J = F ave D t = P f - P To find a , first find the velocity the ball has at the instant the bottom side hits the ground assuming the ball falls 1.50 m. v f 2 = v i 2 + 2 a ( y f - y i ) v f 2 = 0 + 2( - 9.80 m / s 2 )( - 1.50 m - 0) v f = - 5.42 m / s
Solution- Deformed Ball Since this equation used is only valid for constant acceleration this is approximately the average acceleration. The momentum of the ball goes from P i = -m(5.42 m/s) to 0 when it is being deformed. Using the impulse theorem: We can use the same equation to find the acceleration, assumed constant, as the ball moves from where it first touched the ground until it stopped.

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