A mass 1.3 kg is initially at rest at the top of a 1.8 meter high ramp? It slides down the friction-less ramp and collides elastically with an unknown mass which is initially at rest. After colliding with the unknown mass, the 1.3 kg mass recoils and achieves a maximum height (altitude) of only 0.4 m going back up the friction-less ramp. A.Considering the energy of the 1.3 kg mass just before and just after the elastic collision, how much energy is lost by the 1.3 kg mass? B.What is the speed of the 1.3 kg mass just before the elastic collision? C. What is the speed of the 1.3 kg mass just after the elastic collision? D. Considering conservation of momentum, what is the momentum of the unknown mass just after the elastic collision? E. What is the kinetic energy of the unknown mass after the elastic collision? F. What is the velocity of the unknown mass just after the elastic collision? G. What is the mass of the unknown mass in kg?
Answers (1)
It is all about the conversion of momentum. I will set the object that has mass 1.3kg to object 1 and set the object below to object 2. So, first, I calculate the energy of object 1, which is PE(potential energy at the beginning)= KE(kinetic energy at the end) =(1.3kg)*(9.81m/s^2)*(1.8m) =23J. Then the speed, 1/2mv^2=mgh v=square root(2gh) =square root((2)*(9.81m/s^2)*(1.8m)=5.94m/s. The energy of object 2= KE(at the end of object 1) - KE (2nd bounce)= 23J - (1.3kg)*(9.81m/s^2)*(0.4m)= 17.9J. The speed of object 1 when it bounces can use 2ay=Vf^2 - Vo^2 2(-9.81m/s^2)(0.4m)=-Vo^2 Vo= 2.8m/s. You might start recording these numbers on a piece of paper because now is the tricky part. As we know that energy can't be created or destroyed. So, p(momentum of object 1)= p(momentum of object 2). Then (1.3kg)(5.94m/s)- (2.8m/s)(1.3kg)= mv= 4.08kg*m. 1/2mv^2=17.9J mv^2=35.8J v=35.8J/4.08kg*m=8.77m/s and mass=0.465kg. You can ask me question if you want. i will be here.