Answer:
The magnitude of displacement is 0.082m
Explanation:
While the ball is in motion,we have MV + mv= 0 ...eq1
Where M = combined mass of the platform and the two people.
V = velocity of the platform
m = mass of the ball
v = velocity of the ball
The distance that the platform moves is given by:
X = Vt ...eq2
Where t is the time that the ball is in the air.
The time the ball is in the air is given by:
L/(v-V) ...eq3
Where L is the length of the platform
The quantity(v-V) = velocity of the ball relative to the platform.
Combining eq2 and eq3
X = (V/(v - V))L
From eq1 , the ratios of the velocities is V/v = -m/M
X = (V/v)L / (1 - (V/v) = (-m/M) L /(1+ (m/M))
X = -mL/(M + m)
X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)
X = - 10.48/ 122.36
X = -0.082m
The minus sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.
Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.
