Equation of motion, for a simplicity assume that the cart begins to moves at t=0 and reaches a speed of 0.44m/s at t=1.8 sec. Wr
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Refer to the diagram shown below.
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s