Question

A 195-kg projectile, fired with a speed of 116 m/s at a 60.0° angle, breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally. Determine the velocity of the third fragment immediately after the explosion.

Accepted Answer

Step 1: Analyze the problem and identify the key principles. The projectile breaks into three equal fragments at the highest point of its arc. At this point, the velocity of the projectile is horizontal, and the total momentum of the system is conserved in both the horizontal and vertical directions. Additionally, the energy released in the explosion can be determined by comparing the kinetic energy before and after the explosion. Step 2: Calculate the horizontal velocity of the projectile at the highest point. The horizontal velocity remains constant throughout the motion and is given by the horizontal component of the initial velocity: $$ v_x = v_0 \cos(	heta) $$, where $$ v_0 = 116 \; 	ext{m/s} $$ and $$ 	heta = 60.0^\circ . $$Step 3: Apply the principle of conservation of momentum in the horizontal direction. Before the explosion, the total horizontal momentum is $$ p_x = m v_x $$, where $$ m = 195 \; 	ext{kg} . $$After the explosion, the total horizontal momentum is the sum of the horizontal momenta of the three fragments. Two fragments have known velocities, and the velocity of the third fragment can be determined by solving for the unknown horizontal momentum contribution. Step 4: Apply the principle of conservation of momentum in the vertical direction. Before the explosion, the vertical momentum is zero because the velocity is purely horizontal. After the explosion, the vertical momentum of the system must also sum to zero. Use this condition to solve for the vertical velocity component of the third fragment, given that one fragment moves vertically downward with a known speed. Step 5: Calculate the energy released in the explosion. The total kinetic energy before the explosion is $$ KE_{	ext{before}} = \frac{1}{2} m v_x^2 . $$After the explosion, calculate the total kinetic energy of the three fragments using $$ KE_{	ext{after}} = \sum \frac{1}{2} m_i v_i^2 $$, where $$ m_i $$ and $$ v_i $$ are the masses and speeds of the fragments. The energy released is the difference: $$ \Delta E = KE_{	ext{after}} - KE_{	ext{before}} .$$

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