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emmasim [6.3K]
3 years ago
5

A wagon is rolling forward on level ground. friction is negligible. the person sitting in the wagon is holding a rock. the total

mass of the wagon, rider, and rock is 95.5 kg. the mass of the rock is 0.290 kg. initially the wagon is rolling forward at a speed of 0.510 m/s. then the person throws the rock with a speed of 17.5 m/s. both speeds are relative to the ground. find the speed of the wagon after the rock is thrown directly forward.
Physics
1 answer:
vladimir2022 [97]3 years ago
7 0

use the equation mv+mv=mv+mv

95.5(.51)= 95.2(X)+ 17.5(.290)

Now solve for X and that should be your answer

OR

you know that if you throw the rock forward then the the same amount of force is applied backwards onto the person throwing the rock so if the force applied to the rock is mass times velocity then find the force for the cart and subtract the carts force forward by the force the rock exerts backwards.

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Answer:

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c) 5246745J

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Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

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1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

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