Question

A 27.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00-kg steel dart that embeds itself in the lead sphere.

What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Express your answer with the appropriate units.

Answer 1

Since the sphere is in vertical motion therefore to complete the circle the speed at the top (point A) must be
mV² /R = mg
Hence the minimum speed at the top to complete circle
v = (Rg)^1/2 = (2.8*9.81)^1/2 = 5.24 m/s
Now the sphere will move to the bottom of the circle (point B)
Hence due to energy conservation
Initial energy = Final energy
(1/2)mV² + mgH = (1/2)mu²
where u is the speed of the sphere and dart(embedded in spher) at point B
And H is the height of the point A from point B
(1/2)*m*5.24²+ m*9.81*(2*2.8) = (1/2)m*(u²)
u = 11.719 m/s
Since the collision is takes place at point B therefore the
Conservation of momentum takes place
MV = (m+M)u
where M is the mass of the steel dart and V is the speed of dart , and uis the final speed of the sphere and dart embedded in it after collision
5 V= (27+5)11.719
V = 75 m/s