Asked by Van
A ball A having a speed of 44.0 ms-1
collides with another ball, B, of the same mass
which is initially at rest. After the collision A is deflected, 45.0 from its initial direction and
B moves off at an angle of 30.0° to the original direction of A.
a. What is the nature of the collision? (1 Mark)
b. What is the speed of each ball after the collision?
collides with another ball, B, of the same mass
which is initially at rest. After the collision A is deflected, 45.0 from its initial direction and
B moves off at an angle of 30.0° to the original direction of A.
a. What is the nature of the collision? (1 Mark)
b. What is the speed of each ball after the collision?
Answers
Answered by
Henry
M*V1 + M*V2 = M*V3[45o] + M*V4[30o].
V1 + 0 = V3[45o] + V4[30o].
44 = V3*Cos45 + V4*Cos30.
Eq1: 0.707V3 + 0.866V4 = 44,
V3 = 62.2-1.22V4.
V3*sin45 + V4*sin30 = 0.
Eq2: 0.707V3 + 0.5V4 = 0,
1.414V3 + V4 = 0,
V4 = -1.414V3.
V3 = 62.2 - 1.22*(-1.414V3),
V3 = 62.2 + 1.73V3, V3 = -85.2 m/s. = Velocity of ball A.
V4 = -1.414V3 = -1.414*(-85.2) = 120.5 m/s.= Speed of B.
V1 + 0 = V3[45o] + V4[30o].
44 = V3*Cos45 + V4*Cos30.
Eq1: 0.707V3 + 0.866V4 = 44,
V3 = 62.2-1.22V4.
V3*sin45 + V4*sin30 = 0.
Eq2: 0.707V3 + 0.5V4 = 0,
1.414V3 + V4 = 0,
V4 = -1.414V3.
V3 = 62.2 - 1.22*(-1.414V3),
V3 = 62.2 + 1.73V3, V3 = -85.2 m/s. = Velocity of ball A.
V4 = -1.414V3 = -1.414*(-85.2) = 120.5 m/s.= Speed of B.
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