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dangina [55]
3 years ago
12

A 25-kg child runs at 4.0 m/s and jumps onto a shopping cart and holds on for dear life. The cart has mass 15 kg. Assuming the c

hild rides on the cart after the collision, the speed of the child and shopping cart just after the child jumps on is:_______.
a) zero
b) 2.0 m/s
c) 3.0 m/s
d) 2.5 m/s
Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Linear momentum (mass x speed) has to be conserved.

-- Momentum before the jump:

(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s

(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero

Total momentum before the jump:  (100 kg-m/s) + (zero) = (100 kg-m/s)

-- Momentum after the jump:

(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)

-- Momentum after the jump = momentum before the jump

(40 kg) x (speed) = 100 kg-m/s

Divide each side by  40 kg:

Speed = (100 kg-m/s) / (40 kg)

<em>Speed = 2.5 m/s</em>  (d)

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The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

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\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

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Explanation:

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