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Reil [10]
3 years ago
10

When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart

Physics
2 answers:
Karolina [17]3 years ago
7 0

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

I=\Delta p=m\Delta v

where

m = 5.0 kg is the mass of the cart

\Delta v=2.2 m/s is its change in speed

Substituting numbers into the equation, we find

I=(5.0kg)(2.2 m/s)=11 kg m/s

ivolga24 [154]3 years ago
7 0

Solution:

In this question we have given

mass of Cart=5Kg

change in speed,\Delta v=2.2\frac{m}{s}

We have to find, Impulse, I=?

Impulse :

Impulse of a body is equla to its change in momentum.

Therefore,

Impulse of cart is equal to its change in momentum:

I=\Delta p\\I=m\Delta v................(1)

Put values of \Delta v, and m in eq(1)

I=(5.0kg)(2.2\frac{m}{s} )\\I=11 kg\frac{m}{s}

Therefore, the impulse of the cart,I=11 kg\frac{m}{s}

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<h3>Elastic Collision</h3>

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