Question

At an Independence Day celebration show, a large firework is shot from the ground and, in the first stage of its bursting pattern, splits into two unequal parts.

Assume that the firework has a mass of m_o = 0.30 kg and is launched from a cannon at an angle of theta_o= 48.8° with an initial velocity of v= 20 m/s. Just as the firework reaches its hight point in its trajectory, it explodes into two pieces. Immediately after the explosion, the first piece, with a mass of m1 = 0.2 kg flies off at an angle theta_1 = 150° relative to the positive (forward) horizontal axis at a velocity of v1= 5 m/s relative to the ground. Calculate the trajectory angle, theta_2, of the second piece immediately after the explosion. Assume the second piece has a mass of m2 = 0.1 kg, neglect the effect of air resistance, and report your result as an angle relative to the positive (forward) horizontal axis.

(408553) {2D Inelastic Collisions} At an Independence Day celebration show, a large firework is shot from the ground and, in the first stage of its bursting pattern, splits into two unequal parts. Image size: SML Max .--- > AX Part a (1 points) Assume that the firework has a mass of mo = 0.30 kg and is launched from a cannon at an angle of 60 = 48.8° with an initial velocity of Vo = 20 m/s. Just as the firework reaches its hight point in its trajectory, it explodes into two pieces. Immediately after the explosion, the first piece, with a mass of mı = 0.2 kg flies off at an angle 01 = 150° relative to the positive (forward) horizontal axis at a velocity of |01| = 5 m/s relative to the ground. Calculate the trajectory angle, 12 of the second piece immediately after the explosion. Assume the second piece has a mass of m2 = 0.1 kg, neglect the effect of air resistance, and report your result as an angle relative to the positive (forward) horizontal axis. Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 106, 5.23e-8 Enter answer here degrees Your -9.2 degrees Answer

Answer 1

* During explosion 0,=150 30=0 roz? for piece contal component q velocity V -V, COS 30°= -5xC0530 I 2-4-33 mis vertical component of seus of velocity velocity vy, a Vy = V, sin 30 = 5xsinzo for piece ? horizontal component of velocity V = V2 Cos 02 vertical component of velocity =h sino ya ² ... for Horizontal motion , According to law gi conservation momentum ? Initial momentum = final momentum 2. moro cos(48.8) = m, vx, + m₂ x 10.3020% 0.658 = 0.2*(-4.33) + 0.1 % V2cusez Y cos Oz = 48.14 0 for vertical motion, According to law of conservation of momentum :. Initial momentum = final momentum i o m , Vy, 4 m ryn, io = 0.2x2.5 + Ons x Gvz sinon) - 2 sin = 5 Now Divide @by usinor. 5 ton as you 10 = 5.92907 U USO2 trajectory angle a piece.2 after explosion is 5.9290