(a) The speed of the person just before the collision with the object. (b) The t
ID: 3163197 • Letter: #
Question
(a) The speed of the person just before the collision with the object. (b) The tension in the rope just before the collision with the object. (c) The speed of the person and object just after the collision. points) ml 900 m2 Lake A rope of length L is attached to a support at point C. A person of mass mi sits on a ledge at position A holding the other end of the rope so that it is horizontal and taut, as shown. The person then drops off the ledge and swings down on the rope toward position B on a lower ledge where an object of mass m2 is at rest. At position B the person grabs hold of the object and simultaneously lets go of the rope. The person and object then land together in the lake at point D, which is a vertical distance L below position B. Air resistance and the mass of the rope are negligible. Derive expressions for each of the following in terms of m1, m2, L, and g.Explanation / Answer
a) Height desended by the person = L
Potential energy lost = m1Lg
= m1 u2 /2 , kinetic energy gained by the person
u = sqrt(2Lg)
b) Wehn the person reached B the centrifual force m1u2/L acts downard away from the center C and also the gavitation force m1g
Hence tension in the rope T = m1u2/L + m1 g = m1( 2Lg/L +g) = 3m1g
c) let v be the speed of the person and the object just after collision
momentum is conserved m1u = (m1 +m2) v
v = m1u/(m1+m2) = m1(2Lg)1/2 /(m1+m2)
The speed v is in horizontal direction at the point of B
d) height of descent = L , inital verticla velocity vy = 0
L = gt2/2
time of flight t = sqrt(2L/g)
Horizonatl disatnce traveled from point B = vt
= {m1(2Lg)1/2 /(m1+m2) } * sqrt(2L/g)
= 2m1Lg/(m1+m2)
distance from A = L + 2m1Lg/(m1+m2)
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