Suppose that two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. What is the acceleration of the heavier cart relative to that of the lighter cart as they start to move apart?

Solution 23E Introduction According to Newton’s 3rd law of motion, for every action, there is an equal and opposite reaction. According to Newton’s 2nd law of motion, the rate of change of momentum is directly proportional force applied. Therefore, force, = ma Where, m - mass of the body a - acceleration Step 1 Consider, the system is at rest initially. Hence the forces are balanced on each cart. So, consider the force of cart with mass “m” is F. The force, F = ma. Where, ‘a’ is the acceleration of the cart. Since the system is at rest, the force exerted by the cart with mass ‘2m’ also will be same but opposite in direction(balanced force). So, F = 2mx But we know that, F = ma from cart 1. Therefore, ma = 2mx The “m”s on both sides will cancel each other. That is, the acceleration of cart 2 will be, x = a Or x = (a/2) m/s This is the acceleration during the compression of the spring. If the spring is released, the carts will get the same acceleration during compression. Conclusion The acceleration of the heavier cart will be the half of that of the lighter cart and it will be opposite to that of lighter cart.