I din\'t understand the solution in the text help for homework, but this is what

Question

I din't understand the solution in the text help for homework, but this is what I have. Can someone please help and explain
34.) Multiple Concept Example 9 reviews the concepts that play roles in this problem. The drawing in the text shows a collision between two pucks on an air hockey table. Puck A has a mass of 0.025 kg and is moving along the x axis with a velocity of + 5.5 m/s. It makes a collision with Puck B, which has a mass of 0.050 kg and is initially at rest (0.0 m/s). The collision is not head on. After the collision, the two pucks fly apart with the angles shown in the drawing of Puck A is 65o and Puck B is 37o. Find the final speed of a.) Puck A and b.) Puck B.
Section 7.3 Collision in One Dimension
Section 7.4 Collision of Two Dimensions
m1vg1x + m2vf2x = m1vo1x + m2v02x
m1vg1y + m2vf2y = m1vo1y + m2v02y
0.025 kg(5.5 m/s) + 0.050 kg (0.0 m/s) = .1375 + 0 = .1375

m1 = 0.025 kg
v01 = 5.5 m/s
? = 65o
m2 = 0.050 kg
v02 = 0 m/s
? = 37o

Vf1 = v vf1x2 ++ vf2y2
? + tan-1 (vf1y/ vf1x)
m1(vf1x )+ m2(vf2) cos 37o
0.025 kg(5.5 m/s) + 0.050 kg(vf2) cos 37o = 0.1375 + 0.050 kg(vf2)0.7986 = 0.1375 + 0.03993(vf2)
(m1)(v01) sin 65o + m2(v02) =
0.025 kg(5.5 m/s) sin 65o + 0.050 kg( 0 m/s) sin 65o + (0.050 kg)(0 m/s) = 0.1246 + 0 = 0.1246
(m1)(vf1y) +( m2)[(-vf2) sin 37o =
0.025 kg-(5.5 m/s) cos 65o + 0 = 0.1246

Explanation / Answer

mass of puck A is mA = 0.025 kg mass of puck B is mB = 0.050 kg The speed of puck A is uA = 5.5 m/s along X axis The angles made by the puck A after the collision with the horizontal is A = 65o The angles made by the puck B after the collision with the horizontal is B = 37o Let vA and vB be the velocities of puck A and B respectively after collision According to the conservation of linear momentum in X direction mAuA + mB(0) = mAvAcosA + mBvBcosB                 mAuA = mAvAcosA + mBvBcosB According to the conservation of linear momentum in Y direction mA(0) + mB(0) = mAvAsinA - mBvBsinB                         0 = mAvAsinA - mBvBsinB       mAvAsinA = mBvBsinB (0.025 kg)(vA)sin65 = (0.050 kg)(vB)sin37                        vA = 1.328vB Therefore                       mAuA = mAvAcosA + mBvBcosB (0.025 kg)(5.5 m/s) = (0.025 kg)vAcos65 + (0.050 kg)vBcos37                      0.1375 = 0.0106vA + 0.0399vB                      0.1375 = 0.0.106(1.328vB) + 0.0399vB                      0.1375 = 0.0141vB + 0.0399vB                      0.1375 = 0.054vB                             vB = 2.55 m/s                             vA = 1.328(2.55 m/s)                             vA = 3.39 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.