Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a h

Question

Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. The two people, the platform and a 3.36-kg ball are all initially at rest. One person throws the 3.36-kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 119 kg. Because of the throw, this 119-kg mass recoils. How far does the platform move before coming to rest again

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3 years 2021-08-17T04:50:13+00:00 2 Answers 4 views 0

Answers ( )

1. Answer:

Displacement = 0.086m

Explanation:

Given

M = Combined mass of the two people and the platform = 119kg

m = mass of the ball = 3.36kg

v = velocity of the ball

V = Recoil velocity of the platform

t = time spent by the ball in the air

L = length of the platform = 3.12m

MV + mv = 0

The distance moved by the platform is given by; x = Vt

Where t = L/(v – V)

v – V represents the velocity of the ball relative to the platform.

Substitute L/(v – V) for t

so, x = VL/(v-V)

From (MV + mv = 0), we have

MV = -mv

V/v = -m/M

So, x = VL/(v-V) becomes

x = ((V/v)L)/(1 – V/v)

x = ((-m/M)L)/(1 + m/M)

x = -mL/(M+m)

x = -(3.36 * 3.12)/(3.36 + 119)

x = -0.08567505720823798627002

x = -0.086m

The negative sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

The distance moved by the platform before coming to rest is 0.086m

2. 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.