Question #12df8

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Question #12df8

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Answer 1 · 2017-01-06T11:24:04

I got: $v = 6.8 \frac{m}{s}$ $v=6.8m/s$

Explanation:

I would use conservation of momentum $p$ $p$ applied to the entire system of crane and cart and write:
$p_{before} = p_{after}$ $p_("before")=p_("after")$
in numbers:
$1200 \cdot 10.8 = (1200 + 691) v$ $1200*10.8=(1200+691)v$
where on the right you have tye total mass of cart plus load and $v$ $v$ will be the speed after the loading.
we get:
$v = \frac{1200 \cdot 10.8}{1200 + 691} = 6.8 \frac{m}{s}$ $v=(1200*10.8)/(1200+691)=6.8m/s$

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Question #12df8

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