Two identical masses, m, can glide along a frictionless surface, as shown above.

Question

Two identical masses, m, can glide along a frictionless surface, as shown above. The left-hand mass is pushed against a spring-plunger with spring constant k, and the plunger is compressed through a distance 2d. Conversely, the right-hand mass is pushed against a spring-plunger with stronger spring constant 2k, but it is only compressed through a shorter distance d. Both masses are released from rest at the same time. When they eventually collide, they stick together.

A. Find an expression for the final velocity of the stuck-together pair of masses after collision. Specifiy whether it is to the right or to the left. Express your final answer only in terms of m,k,d, and any necessary mathematical constants.

B. Find the fraction of initial mechanical energy that is lost in the collision.

Instead of sticking together, suppose that the two masses collide perfectly elastically.

C. For this new perfectly elastic collision, the total kinetic energy (Ktot) of all masses immediately before collision is ______ the Ktot of all masses immediately after collision.

A. less than

B. the same as

C. greater than

D. In which case is the final total kinetic energy of all masses immediately after collision (Ktot, final) greater?

A. stuck-together collision

B. perfectly elastic collision

Explanation / Answer

velocity obtained by the left hand mass = (k/m)^.5*2d velocity obtained by the right hand mass = (2k/m)^.5*d after collision momentum is conserved m( (k/m)^.5*2d- (2k/m)^.5*d) = 2m v v = (k/m)^.5*(2-1.414)/2*d = .3*(k/m)^.5*d towards right final mechanical energy = .5*2m*v^2 = .5*2*m*k/m*.09 = .09k*d^2 initial mechanical energy = .5*k*4*d^2+.5*k*d^2 = 3k*d^2 fraction of energy lost = (3-.09)k*d^2/(3k*d^2) = .97 = 97% C) for perfectly elastic collision energy is conserved so answer is B. the same as D) perfectly elastic collision. in the other case energy is lost as heat its not conserved .

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