1) Two blocks are free to slide along the frictionless wooden track shown below.
ID: 1796517 • Letter: 1
Question
1) Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.90 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.: m
4. -1 points SerPSE8 9.P027.WI. My Notes Ask Your Teach Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 4.90 kg is released from the position shown, at height h 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision 1n m2 Need Help? ReadiltWatch t Talk to a Tutor Read It Watch It 5. -12 points SerPSE8 9.P.028 My Notes Ask Your Teach Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.5 m/s toward the east and the other is traveling north with speed v2i Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 58.0° north of east. Determine the initial speed v2/ of the northward moving vehicle m/s The speed limit for both roads is 35 mi/h and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth? Yes O NoExplanation / Answer
PE1
= mgh
= 4.9*9.8*5.0 = 240.1 J
V1 = (2gh) = 9.899 m/s
Use momentum to find V2:
V2 = m1*V1/m2 = 4.9*9.899/10.5 = 4.6195 m/s
KE2 = ½m2*V2² = 112.035 J
PE1' = PE1 - KE2 = 240.1-112.035 J
=128.065J
h' = PE1'/(m1*g) = 128.065/(4.9*9.8) = 2.6669 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.