Dash is standing on his frictionless skateboard with three balls, each with a mass of \(100 \mathrm{g}\), in his hands. The combined mass of Dash and his skateboard is \(60 \mathrm{kg} .\) How fast should dash throw the balls forward if he wants to move backward with a speed of \(0.50 \mathrm{m} / \mathrm{s} ?\) Do you think Dash can succeed? Explain.

To solve this problem, we need to first analyze the situation before and after Dash throws the balls. Using the law of conservation of momentum, we can determine the speed at which he should throw the balls.

Before Dash throws any balls, neither he nor the skateboard nor the balls are moving, so their initial momentum is 0.

After Dash throws the balls, the total momentum in the final state will be equal to the momentum of the three balls and the momentum of Dash and his skateboard. Let the speed at which Dash throws the balls be v (in m/s). The momentum for each ball would be equal to the product of mass and velocity, which is \(0.1\mathrm{kg} \times v\). Since there are three balls, the total momentum of the balls will be \(3 \times (0.1\mathrm{kg} \times v) = 0.3\mathrm{kg} \times v\). For Dash and his skateboard, their final momentum will be equal to their combined mass (60 kg) multiplied by their final velocity (-0.50 m/s), since they are moving in the opposite direction: \((60\mathrm{kg})(-0.50\mathrm{m/s}) = -30\mathrm{kg}\mathrm{m/s}\).

Using the conservation of momentum principle, the total initial momentum must equal the total final momentum. Since the initial momentum is 0, the sum of the final momenta should also be 0: \(0.3\mathrm{kg} \times v - 30\mathrm{kg}\mathrm{m/s} = 0\)

Now solve the equation for v: \(v = \frac{30\mathrm{kg}\mathrm{m/s}}{0.3\mathrm{kg}} = 100\mathrm{m/s}\)

Dash needs to throw the balls with a speed of 100 m/s to move backward with a speed of 0.50 m/s. In reality, it would be difficult for Dash to throw the balls at such a high speed, and so practically speaking, it's unlikely that he could achieve the desired result.