Question

A projectile of mass 2.75 kg is fired in the air at an angle of 45.0^\circ45.0 ∘ to the horizon at a speed of 290.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 2.00 kg, 0.70 kg, and 0.05 kg. The 2.00 kg part falls straight down after breakup with an initial speed of 10.0 m/s, the 0.70 kg part moves in the original forward direction, and the 0.05 kg part goes straight up.Find the speeds of the 0.7 and 0.05 kg pieces immediately after the break-up

Answer 1

Answer:

Immediately after the breakup, the 0.05 kg piece moves at 0m/s and the 0.7 kg piece at 1139.29 m/s.

Explanation:

This problem can be treated via momentum conservation. Initially, the projectile moves in parabolic movement. At its highest point, the velocity vector has only the horizontal component, $v_x=v_i\cos45^{\circ}$ which is, with $v_i=290\text{m/s}$,$v_x\approx205,06\text{m/s}$. Therefore, when it breaks down, the y component of the velocity is 0, so the momentum of the 2kg and 0.05 kg parts must sum 0. Since the 2kg goes straight down and the 0.05 kg goes straight up, immediately after the breakup non of them have had time to accelerate (change their velocities) so both equal 0. On the other hand, all the momentum of the original projectile is transferred to the 0.7 kg part. So, applying the conservation of momentum equation,

$Mv=m_{0.7}v_{0.7}$,

being M and v the original projectile's mass and velocity at the highest point and $m_{0.7}\,\text{and}\,v_{0.7}$ the mass and velocity of the 0.7 kg part. Plugging all the numbers,

$2.75\times290 \text{kgm/s}=0.7\text{kg}\,v_{0.7}\Rightarrow v_{0.7}=\dfrac{797.5}{0.7}\text{m/s}=1139.29\,\text{m/s}.$