A 150 g air glider has a velocity of 0.5 m/s on an air track until acted upon by

Question

A 150 g air glider has a velocity of 0.5 m/s on an air track until acted upon by the force that is plotted above.

1. Calculate the impulse acting on the cart.

2. Calculate the new velocity of the cart after the impulse.

3. Suppose the 150 g glider, now moving at the speed you just calculated, strikes a motionless 250 g glider and the two stick together. How fast is the combination moving immediately after the collision?

4. If the collision lasted 8.7 ms, what average force acted on the two carts? (Note: This collision is not shown in the diagram.)

Explanation / Answer

   ( J (impulse) = Delta mv = int_{}^{} F dt ) Impulse is defined as a change in momentum of body.

1. For this problem impulse will be aread under this curve which is area of this triangle.

= 1/2 * 10 * 60 = 300 N-ms = 0.3 N-s

2. SInce we know impulse which is equal to change in momentum,thus

= 0.3N-s = m(Vfinal - Vinitial) = 150*10^-3(Vfinal - 0.5)

Vfinal = 2.5 m/s

3. After Collision, since no external force is present,momentum of system will be preserved by law of conservation of momentum.vinitial = 0 for 250 gm glider

weight of combination = 400gm

initial momentum = 150gm *2.5m/s + 250gm * 0m/s(other glider is at rest)

final momentum = 400gm * V

400gm * V = 150gm *2.5m/s

V = 75/80 m/s = 0.9375 m/s

4. Let average force be F.Since initial velocity is zero for second glider so is momentum,

From equation given on top F*delta t = delta mv = 250gm * 0.9375m/s(This impulse will change the momentum of second body)

Since t = 8.7ms so ,

F = 250gm * 0.9375/8.7ms = 26.94 N(average force)

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