A particle P of mass m is fastened to the end of a string with length L. The par

Question

A particle P of mass m is fastened to the end of a string with length L. The particle starts from rest, in a position where the string is taut, and the direction of the string forms an angle of 60 degrees with the vertical. In the lowest point of the circular arc, the particle P hits the body A. Before the collision the body A is at rest on a frictionless horizontal table. The body A has a mass nm where n is some positive number. The central collision between P and A is completely elastic. The acceleration of gravity is g, and we neglect air resistance. The known quantities are thus n, m, L, and g.

1) Just before the collision the particle P has a kinetic energy T_0. Find T_0.

2) Determine the velocity of the body A and the particle P just after the collision. Furthermore, determine that fraction q of the kinetic energy T_0 of the particle that is transferred to the body A.

Explanation / Answer

mgl*(1-cos)=1/2mv^2

so K.e=mgl*(1-cos) v=(2gl(1-cos)

2)m(2gl(1-cos)=mV'+nm*V''

e=1=v''-v'/V==>v''-v'=(2gl(1-cos)

take (2gl(1-cos)= Z

V''=Z+V'==>V'+n(Z+V')

V'=Z(1-n)/(n+1)

V''=Z+V'= 2Z/1+n wherr (2gl(1-cos)= Z

fraction of K,e transsfered= K.eInitial of 1-keFinal of 1/k.e1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.