A projectile of mass m on a path with no air resistance breaks into two pieces o

Question

A projectile of mass m on a path with no air resistance breaks into two pieces of equal mass m/2. Its velocity vector in m/s at t = 0 at the cartesian position r vector = (0, 0) is described by: V_i vector= 10i^+ 30j^, m/s, where i^is a unit vector parallel to the ground, and j^is a unit vector in the vertical direction. The firework explodes into two pieces immediately after 4 s. a. Calculate its position r vector and velocity v vector at t = 4s. b. Draw a diagram illustrating the trajectory. c. Calculate the linear momentum at t = 4 s, and express in vector form. d. At t = 4 s, one piece begins a new trajectory after the explosion: V_A vector = 20 i^+ 5j^m/s. Using the fact that linear momentum is always conserved, calculate the velocity of the second piece v_B vector at t = 4 s. c. Where does each piece land?

Explanation / Answer

Solution:

The velocity of particle will be constant as 10 in x direction as there is no force in that direction, but its velocity changes in y direction due to gravity.

So , after 4 sec, velocity in x direction = 10i

velocity in y direction vy = 30 - 4X10 = -10

So, velocity vector after 4 seconds, V = 10i - 10j

Position after 4 seconds:

Chnage in position in x direction = 10x4 = 40

Change in position in y direction = 30x4 - 1/2x10x4x4 = 120 - 80 = 40

So position vector after 4 seconds R = 40 i + 40j

Linear momentum at t=4 sec, P = m(10 i -10j)

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