Two forces are applied to a tree stump to pull it out of the ground. Force A has a magnitude of 2410 newtons (N) and points 36.0
° south of east, while force B has a magnitude of 4470 N and points due south. Using the component method, find the (a) magnitude and (b) direction of the resultant force A + B that is applied to the stump. Specify the direction as a positive angle with respect to due east.
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Answer:
(a). The speed of the first ball after the collision is 1.95 m/s.
(b). The direction of the first ball after the collision is 44.16° due south of east.
Explanation:
Given that,
Velocity of one ball u₁= 2.2i m/s
Velocity of second ball u₂=- 0.80i m/s
Final velocity of the second ball v₂= 1.36j m/s
The mass of the identical balls are
(a). We need to calculate the speed of the first ball after the collision
Using law of conservation of momentum
Along X- axis
Put the value into the formula
Along Y-axis
Put the value into the formula
Then the final speed of the first ball
(b) We need to calculate the direction of the first ball after the collision
Using formula of direction
Negative sign shows the direction of first ball .
Hence, (a). The speed of the first ball after the collision is 1.95 m/s.
(b). The direction of the first ball after the collision is 44.16° due south of east.