Two blocks slide on a collision course across a frictionless surface, as in the

Question

Two blocks slide on a collision course across a frictionless surface, as in the figure. The resulting collision is inelastic. The first block has mass

M = 1.20 kg

and is initially sliding due north at a speed of

Vi = 9.15 m/s.

The second block has mass

m = 7.00 102 kg

and is initially sliding at a speed of

vi = 13.5 m/s

directed at an angle

= 27.5°

south of east. Immediately after the inelastic collision, the second block is observed sliding at

vf = 4.75 m/s

in a direction of

= 21.5°

north of east.

Determine the components of the first block's velocity after the collision.

Explanation / Answer

m*o + m(13.5)*costheta = m*(4.75*cosphi) + M*Vxf

m*(13.5*costheta - 4.75*cosphi) = M*vf

Vxf = 0.44 m/s

M*9,15- m(13.5)*sintheta = m*(4.75*sinphi) + M*Vyf

Vyf = 8.68 m/s

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