A sphere of mass M1-2.12 kg tied to a string of negligible mass is released from

Question

A sphere of mass M1-2.12 kg tied to a string of negligible mass is released from rest, then it hit a sphere of mass M2 is initially at rest. The collision is elastic. If M1 is released when 33.8° and the length of the string is L 3.44 m. 6.92 kg, which M2 a) Determine the speed of M1 just before the collision 3.376 m/s You are correct. Your receipt no. is 167-5152 b) Determine the speed of M1 just after the collision -6.92 m/s Submit Answer Incorrect. Tries 4/10 Previous Tries c) Determine until what height ill rise up M2 after the collision Submit Answer Tries 0/10

Explanation / Answer

Given

M1 = 2.12kg,M2 = 6.92 kg

theta = 33.8 degrees

L = 3.44 m

when M1 is released from a height h = L(1- cos theta )

h = 3.44(1- cos 33.8) m = 0.5814134 m

by conservation of energy

mgh = 0.5*mv^2 before colliding M2

M1*g*h = 0.5*M1*v^2

V = sqrt(2*g*h)

V = sqrt(2*9.8*0.5814134) m/s = 3.376 m/s

b) the collision is elastic collision

the conservation of momentum the speed of the M1 after the collision is zero imparting the momentum to M2 and reaches to aheight h2

that is 2.12*3.376 = 6.92*V2 ==> V2 = 1.03427 m/s

c)

so that 0.5*M2*V2^2 = M2*g*h2

h2 = 0.5*V2^2 / g

h2 = 0.5*1.03427^2 /9.8 m = 0.054577267 m

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