By accident, a large plate is dropped and breaks into three pieces. The pieces f
ID: 2008559 • Letter: B
Question
By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.95 m/s and v2 = 1.90 m/s. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing, find the masses of pieces 1 and 2.
By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.95 m/s and v2 = 1.90 m/s. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing, find the masses of pieces 1 and 2.Explanation / Answer
mass m3 = 1.3 kg velocity of first piece v1 = 2.95 m/s velocity of second piece v2 = 1.90 m/s velocity of third piece v3 = 3.07 m/s components of velocity of the first piece : v1x = v1 cos65 (-i^) = -1.246 i^ m/s v1y = v1 sin65 (j^) = 2.673 j^ m/s components of velocity of the second piece : v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg velocity of second piece v2 = 1.90 m/s velocity of third piece v3 = 3.07 m/s components of velocity of the first piece : v1x = v1 cos65 (-i^) = -1.246 i^ m/s v1y = v1 sin65 (j^) = 2.673 j^ m/s components of velocity of the second piece : v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg v1x = v1 cos65 (-i^) = -1.246 i^ m/s v1y = v1 sin65 (j^) = 2.673 j^ m/s components of velocity of the second piece : v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg = 2.673 j^ m/s components of velocity of the second piece : v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg components of velocity of the second piece : v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg v2x = v2 cos45 (i^) = 1.343 i^ m/s v2y = v2 sin45 (j^) = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg = 1.343 j^ m/s components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg components of velocity of the third piece : v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kg v3x = 0 m/s = 1.343 i^ m/s v3y = -3.07 j^ m/s ........................................................... from conservation of momentum , the total momentum Px = 0 and Py = 0 Px = m1v1x + m2v2x + m3v3x = 0 m1v1x + m2v2x + m3v3x = 0 (m1)(-1.246 i^) + (m2)(1.343 i^) + (1.3)(0) = 0 -1.246 m1 + 1.343 m2 = 0 m2 = (0.9277) m1 ............ (1) Py = m1v1y + m2v2y + m3v3y = 0 m1v1y + m2v2y + m3v3y = 0 (m1)(2.673 j^) + (m2)(1.343 j^) + (1.3)(-3.07 j^) = 0 2.673 m1 + 1.343 m2 - 3.991 = 0 .......... (2) substitute the eq (1) , in eq (2), we get 2.673 m1 + 1.343 (0.9277) m1 = 3.991 m1 = 1.018 kg this value substitute in eq (1), we get m2 = 0.9447 kgRelated Questions
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