How to find initial speeds of two masses, when the final velocity is given For 2

Question

How to find initial speeds of two masses, when the final velocity is given

For 2a) I got Px=(m1+m2)vx

For 2b) I got Py=(m1+m2)vy

After that, once I got to 2c) I got lost. I would appreciate any help if possible.

As depicted below, two masses (m, and m2) are involved in a Totally Inelastic Collision. The directions of their initial velocities are given, but the magnitudes of their initial velocities (speeds) are not. The direction and magnitude of their shared common velocity is given. An "xy" coordinate system has been set up for you to use in the +y mi = 3.00 kg lal = 53.13° m2 = 4.50 kg l, I = ??.?? m/s m12 330.00 2 2, 1-1 2.00 "/s

Explanation / Answer

Pfx = Pix

(m1+m2)*v12fx = m1*v1ix + m2*v2ix

(m1+m2)*v12f*cos(beta) = m1*1vi*cos(alpha) + m2*v2i*cos0

(3 + 4.5)*12*cos30 = (3*v1i*cos53.13) + 4.5*v2i............(1)

========================

Pfy = Pi

-(m1+m2)*v12fy = -m1*v1iy + m2*v2iy

(m1+m2)*v12f*sin(beta) = -m1*1vi*sin(alpha) + m2*v2i*cos90

(3 + 4.5)*12*sin30 = (3*v1i*sin53.13) ...............(2)

=========================================

from 2

(3 + 4.5)*12*sin30 = (3*v1i*sin53.13)

45 = 2.4*v1i

v1i = 45/2.4

v1i = 18.75 m/s    <<<<================ANSWER

from 1

(3 + 4.5)*12*cos30 = (3*v1i*cos53.13) + 4.5*v2i

(3 + 4.5)*12*cos30 = (3*18.75*cos53.13) + 4.5*v2i

78 = 33.75 + (4.5*v2i)

v2i = 9.83 m/s <<<<================ANSWER

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