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Ball A of mass 1.5 kg swings down from rest on a string of length 0.8m, with a s

ID: 1998876 • Letter: B

Question

Ball A of mass 1.5 kg swings down from rest on a string of length 0.8m, with a starling angle theta = 50 degree. At the lowest point of its swing it undergoes an elastic collision with block B of mass 2.0 kg which is, rest on a friction less surface. What is the speed of block B immediately alter this collision? Block B then undergoes a completely inelastic collision with block C. Block C of mass 2.6 kg is initially at rest and in contact with an unscratched spring with spring constant k. W hat is the speed of blocks B and C immediately after the inelastic collision? The blocks move to the right a distance of 8.6 cm before momentarily coming to rest. What is the spring constant, k? What is the period of the resulting oscillation of the masses? (Assume that ball A no longer interferes with this oscillation.) What maximum acceleration does the pair of blocks, B and C, experience?

Explanation / Answer

A) initail height of ball A above its lowest point,

h = L*(1-cos(theta))

= 0.8*(1 - cos(50))

= 0.28577 m/s

speed of A just before hitting block B,

vA = sqrt(2*g*h)

= sqrt(2*9.8*0.28577)

= 2.37 m/s

speed of block B immdeditely after the collision,

vB = 2*mA*vA/(mA+mB)

= 2*1.5*2.37/(1.5 + 2)

= 2.03 m/s <<<<<<<<<<-------------------Answer

B) let V is the speed of blocks B and C immedeiately after the collsion.

mB*vB = (mB + mC)*V

==> V = mB*vB/(mB + mC)

= 2*2.03/(2 + 2.6)

= 0.883 m/s <<<<<<<<<<-------------------Answer

C) Apply Consrvation of energy

(1/2)*k*x^2 = (1/2)*(mB + mC)*V^2

k = (mB + mC)*V^2/x^2

= (2 + 2.6)*0.883^2/0.086^2

= 485 N/m <<<<<<<<<<-------------------Answer


D) angular frequency of motion, w = sqrt(k/(mB + mC))

= sqrt(485/(2 + 2.6))

= 10.3 rad/s

Time period, T = 2*pi/w

= 2*pi/10.3

= 0.61 s <<<<<<<<<<-------------------Answer

E) a_max = Fmax/(mB + mC)

= k*x/(mB + mC)

= 485*0.086/(2 + 2.6)

= 9.07 m/s^2 <<<<<<<<<<-------------------Answer

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