Question

Two air-track gliders are held together with a string. The mass of glider A is twice that of glider B. A spring is tightly compressed between the gliders. The gliders are initially at rest and the spring is released by burning the string. If glider B has a speed of 3 m/s after the release, how fast will glider A be moving?

Answer 1

Let the mass of glider B be m, so the mass of glider A is 2m. Applying the conservation of momentum, we know that the initial momentum equals the final momentum.

Initial momentum = 0
Final momentum = momentum of glider A + momentum of glider B = 2m * v_A + m * v_B

v_A is the velocity of glider A, and v_B = 3 m/s is the velocity of glider B.

Since the initial momentum is 0, we can write the equation as:

0 = 2m * v_A + m * v_B

Substitute v_B with 3 m/s:

0 = 2m * v_A + m * 3

To isolate v_A, divide the entire equation by m:

0 = 2v_A + 3

Now, we can solve for v_A:

2v_A = -3
v_A = -3/2 = -1.5 m/s

So, glider A will be moving at -1.5 m/s, in the opposite direction of glider B.