A 0.140-kg glider is moving to the right on a frictionless, horizontal air track
ID: 249971 • Letter: A
Question
A 0.140-kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.75 m/s. It has a head-on collision with a 0.280-kg glider that is moving to the left with a speed of 2.14 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic. 0.140-kg glider magnitudedirection
0.280-kg glider magnitude
direction
A 0.140-kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.75 m/s. It has a head-on collision with a 0.280-kg glider that is moving to the left with a speed of 2.14 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic. 0.140-kg glider magnitude
direction
0.280-kg glider magnitude
direction
0.140-kg glider magnitude
direction
0.280-kg glider magnitude
direction
Explanation / Answer
m1= 0.140 Kg; m2= 0.280 Kg; u1=0.75 m/s; u2= 2.14 m/s
by using law of conservation momentum:
m1*u1+m2*u2=m1*v1+m2*v2
(0.14*0.74)-(0.28*2.34)=0.14*v1+0.28*v2
v1+ 2v2=-3.94
v1= -3.94 -2v2 (I)
conserving energy of the system:
1/2*m1*u1^2+1/2*m2*u2^2=1/2*m1*v1^2+1/2*m2*v2^2
m1*u1^2+m2*u2^2=m1*v1^2+m2*v2^2
(0.14*0.742)+(0.28*2.342)=0.14v12+0.28v22
1.609= 0.14v12+0.28v22
we know the equation (I) for v1:
1.609=0.14(-3.94-2v2)2+0.28v22
1.609=0.14 (15.523+15.76v2+4v22)+0.28v2
0.84v22+2.206v2+0.564=0
v2= -0.287 ; v2= -2.33
v1= -3.366; v1= 0.72
so, final answer is
v1= 0.72 m/s towards right
v2= 2.33 m/s towards left
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