A 0 5-kilogram cube slides at 20 m/s on a smooth frictionless surface toward a s
ID: 1523335 • Letter: A
Question
A 0 5-kilogram cube slides at 20 m/s on a smooth frictionless surface toward a stationary sphere, shown below left. The sphere is half the volume the cube, but is eight times as dense. The cube strikes the sphere at time t=0. A plot of the force exerted on the sphere by the cube as a function of time is shown above right. What is the impulse applied to the sphere? What is the change in momentum of the sphere? What is the speed of the sphere immediately following the collision? What is the velocity of the cube immediately following the collision? (State both direction and magnitude.)Explanation / Answer
Apply conservation of momentum to the sphere and cube
m1u1 + m2 u2 = m1v1 + m2 v2
The final speed of the sphere after collisiion is
v2 = 2 m1/ m1 + m2 ) u1 + (m2- m1/ m1 + m2) u2
since u2 = 0
v2 = 2 m1/ m1 + m2 ) u1
=2 (0.5)/(0.5+4(0.5) * 20
=8 m/s
the final velocity of the cube
v1 = ( m1- m2/ m1 + m2) u1
= (0.5-4(0.5)/0.5+4(0.5) * 20
=-12 m/s
(a)
Impulse = change in momentum of the sphere
=m2( v2- u2)
= 4(0.5) (8 m/s - 0)
= 16 kg m/s
(b)
change in momentum of the sphere is
=m2( v2- u2)
= 4(0.5) (8 m/s - 0)
= 16 kg m/s
(c)
v2 = 2 m1/ m1 + m2 ) u1
=2 (0.5)/(0.5+4(0.5) * 20
=8 m/s
(d)
v1 = ( m1- m2/ m1 + m2) u1
= (0.5-4(0.5)/0.5+4(0.5) * 20
=-12 m/s
magnitude of velocity is 12 m/s
toward left side
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.