A block of mass mb=4 kg rests on a frictionless ice sheet, connected to a fixed

Question

A block of mass mb=4 kg rests on a frictionless ice sheet, connected to a fixed post by an ideal spring with spring constant k=76 N/m. The block is initially at rest, with the spring at its rest length of 40 cm. The block is then struck head on by a mp=0.1 kg hockey puck moving at 25 m/s, which rebounds with a speed of 15 m/s. You may assume the collision takes place quickly enough that spring does not compress during the collision. (a)Find the maximum compression of the spring and the amplitude of the resulting oscillations of the block. (b) Find the position of the block (relative to its initial position) at t2=6 seconds later. (c) Find the momentum of the block (relative to its initial position) at t2=6 seconds later **please explain and show work very clearly on how to get to answers, to understand**

Explanation / Answer

T1 = (m1g - m1a) T2 = m2a the torque on the pulley is F*R where F = T1 - T2 --> torque = T1*R - T2*R ---> torque = (m1g - m1a)*R - m2a*R and in addition is torque = I*a = I*a/R with a = a/R since the tangential acceleration of the pulley is equal to the tangential acceleration of the whole system, we can write (m1g - m1a)*R - m2a*R = I*a/R (m1g - m1a)*R^2 - m2a*R^2 = I*a now solve for a m1gR^2 - m1aR^2 - m2aR^2 = I*a a(- m1R^2 - m2R^2 - I) = - m1gR^2 a = m1gR^2/(m1R^2 - m2R^2 - I) --> divide by R^2 a = m1g/(m1 + m2 + I/R^2) put in the given: a = 4.3*9.81/(4.3 + 3 + 10/0.5^2) a = 0.89 m/s^2 so angular acc. = a/r= 0.89/ 0.5 = 1.78 ans

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