2. A projectile fired vertically upwards exploded into two fragments at the top of the trajectory. The velocity,of the first fragment weighing 0.5 kg immediately after the rupture is directed horizon- tally and is equal to 800m/s. The mass of the second fragment is 2 kg . Accept that the resistance of the spirit can be ignored. a) What is the momentum of the first fragment immediately after the rupture? b) What is the velocity of the second fragment immediately after the rupture? c) Which fragment will fall to the ground first?Justify your answer.

a) The momentum of the first fragment immediately after the rupture can be calculated using the formula:<br /><br />\[ \text{momentum} = \text{mass} \times \text{velocity} \]<br /><br />Given that the mass of the first fragment is 0.5 kg and its velocity is 800 m/s, we can substitute these values into the formula:<br /><br />\[ \text{momentum} = 0.5 \, \text{kg} \times 800 \, \text{m/s} = 400 \, \text{kg} \cdot \text{m/s} \]<br /><br />Therefore, the momentum of the first fragment immediately after the rupture is 400 kg·m/s.<br /><br />b) To find the velocity of the second fragment immediately after the rupture, we can use the conservation of momentum. The total momentum before the rupture must be equal to the total momentum after the rupture. Since the projectile was fired vertically upwards and exploded into two fragments at the top of the trajectory, the initial momentum was zero. Therefore, the momentum of the second fragment must be equal and opposite to the momentum of the first fragment.<br /><br />Given that the mass of the second fragment is 2 kg, we can use the formula:<br /><br />\[ \text{momentum} = \text{mass} \times \text{velocity} \]<br /><br />Substituting the mass of the second fragment and the momentum of the first fragment, we get:<br /><br />\[ 400 \, \text{kg} \cdot \text{m/s} = 2 \, \text{kg} \times \text{velocity} \]<br /><br />Solving for velocity, we find:<br /><br />\[ \text{velocity} = \frac{400 \, \text{kg} \cdot \text{m/s}}{2 \, \text{kg}} = 200 \, \text{m/s} \]<br /><br />Therefore, the velocity of the second fragment immediately after the rupture is 200 m/s.<br /><br />c) Both fragments will fall to the ground at the same time. This is because the vertical component of the velocity for both fragments is zero immediately after the rupture. Since there is no vertical force acting on the fragments (assuming the resistance of the air can be ignored), both fragments will continue to fall freely under the influence of gravity. Therefore, they will reach the ground at the same time.