A small block of mass M is released from rest at the top of the curved frictionl

Question

A small block of mass M is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with a speed 3.5v0 when it collides with a larger block of mass 1.5M at rest at the bottom of the incline. The larger block moves to the right at a speed 2v0 immediately after the collision. Express your answers to the following questions in terms of the given quantities and fundamental constants. Determine the height h of the ramp from which the small block was released. Determine the speed of the small block after the collision. The larger block slides a distance D before coming to rest. Determine the value of the coefficient of kinetic friction mu between the larger block and the surface on which it slides. Indicate whether the collision between the two blocks is elastic or inelastic. Justify your answer.

Explanation / Answer

initial potential energy of the block with mass M=M*g*h potential energy of the mass M at the bottom=0 kinetic energy at thebottom=(1/2)*M*(3.5*v0)2=6.125*M*v02 6.125*M*v02=M*g*h h=0.625*v02 v=velocity of the block with mass M after collision M*3.5*v0+1.5*M*0=1.5*M*2*v0+M*v M*3.5*v0=3*M*v0+M*v v=-0.5*v0 =coefficient of friction frictional force=*1.5*M*g work done by friction=*1.5*M*g*D change in kinetic energy of the block with mass1.5*M=(1/2)*1.5*M*4*v02=3*M*v02-0 change in kinetic energy=work done by frictional force 3*M*v02=*1.5*9.8*M*D =(0.204*v02)/D initial kinetic energy of thesystem=6.125*M*v02 final kinetic energy of thesystem=(1/2)*M*(1/2)*v02+(1/2)*1.5*M*4*v02=3.25*M*v02 since kinetic energy is not conserved the collision isinelastic

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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