A 0.50-kg block attached to an ideal spring with a spring constant of 80N/m osci
ID: 1476205 • Letter: A
Question
A 0.50-kg block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50m/s. The greatest speed of the block is: 0.23m/s 0.32m/s 0.55m/s 0.71m/s 0.93m/s Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that of spaceman B is 90 kg. Spaceman A pushes B away from him with B attaining a final speed of 0.5m/s. The final recoil speed of A is: zero 0.38m/s 0.5m/s 0.67m/s l.0m/s A 0.2-kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse on the ball during the collision is: I0N s upward I0N s downward 2.0N s upward 2.0N s downward 9.8N s upwardExplanation / Answer
Problem 6.
mass of block attached is m = 0.5 kg, k = 80 N/m, at dx= 4 cm, v is 0.5 m/s
total mechanical energy of the spring mass system is sum of kinetic energy and elastic potential energy
E = 1/2 mv^2 + 1/2 kX^2
= 1/2 *0.5*0.5^2 + 1/2 *80*16*10^-4
= 0.1265 J
now for maximum speed of the block will be at mean position that only kinetic energy
E = 1/2 mv^2
v = sqrt(2E/m)
= sqrt(2*0.1265/0.5)
V = 0.7113 m/s
Problem 7.
change in momentum mAVA = MBVB
VA = MBVB/MA
= 90*0.5/120
= 0.375 m/s or 0.38 m/s
Problem 8
impulse = change in momentum
= mu - mv
m= 0.2 kg, u = 30 m/s, v= -20 m/s
change in momentum = m(u-(-v))
= m (u+v)
= 0.2(30+20)
= 0.2*50
= 10 N
the direction is upward
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