A glider of mass 0.140 Kg is moving to the right on a frictionless, horizontal air track with a speed of 0.800 m/s . It has a head-on collision with a glider 0.301 KG that is moving to the left with a speed of 2.12 m/s . Suppose the collision is elastic.
How do I find:
Find the magnitude of the final velocity of the 0.140 kg glider.
Find the magnitude of the final velocity of the 0.301 kg glider.
How do I find:
Find the magnitude of the final velocity of the 0.140 kg glider.
Find the magnitude of the final velocity of the 0.301 kg glider.
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momentum and energy are conserved; before collision, the total momentum is (calling motion to the right positive):
0.14kgx0.8m/s - 0.301kg x 2.12 m/s = -0.53 kgm/s
this will be the momentum after collision; if we call vA and vB the velocities of the two pucks after collision, we have that the momentum after collision is
0.14vA + 0.301vB and this will equal -0.53kgm/s
energy conservation tells us the energy before collision is equal to the energy after collision, since energy is a scalar, all quantities are positive and
energy before collision = 1/2 x 0.14kg x (0.8m/s)^2 + 1/2x0.301kg x (2.12m/s)^2
energy before collision = 0.72J
and this equals 1/2x 0.14 vA^2 + 1/2 x 0.301xvB^2
use the momentum equation to write vB as:
vB = (-0.53 - 0.14vA)/0.301
substitute this into the energy equation to write that equation as a quadratic in vA, solve for vA, then solve for vB
I get va=-3.18m/s and vb = -0.28m/s
0.14kgx0.8m/s - 0.301kg x 2.12 m/s = -0.53 kgm/s
this will be the momentum after collision; if we call vA and vB the velocities of the two pucks after collision, we have that the momentum after collision is
0.14vA + 0.301vB and this will equal -0.53kgm/s
energy conservation tells us the energy before collision is equal to the energy after collision, since energy is a scalar, all quantities are positive and
energy before collision = 1/2 x 0.14kg x (0.8m/s)^2 + 1/2x0.301kg x (2.12m/s)^2
energy before collision = 0.72J
and this equals 1/2x 0.14 vA^2 + 1/2 x 0.301xvB^2
use the momentum equation to write vB as:
vB = (-0.53 - 0.14vA)/0.301
substitute this into the energy equation to write that equation as a quadratic in vA, solve for vA, then solve for vB
I get va=-3.18m/s and vb = -0.28m/s
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Glider 1 = 0.140 kg glider
Glider 2 = 0.301 kg glider
The below equations are derived using a combination of conservation of linear momentum, and conservation of energy.
v1f = (m1-m2)/(m1+m2)*v1i
= (0.140 - 0.301)/(0.140+0.301)*0.800 = -0.292 m/s (negative means it reverses direction)
v2f = 2m1/(m1+m2)*v1i = 2*0.140/(0.140+0.301)*0.800 = 0.508 m/s
Glider 2 = 0.301 kg glider
The below equations are derived using a combination of conservation of linear momentum, and conservation of energy.
v1f = (m1-m2)/(m1+m2)*v1i
= (0.140 - 0.301)/(0.140+0.301)*0.800 = -0.292 m/s (negative means it reverses direction)
v2f = 2m1/(m1+m2)*v1i = 2*0.140/(0.140+0.301)*0.800 = 0.508 m/s