Consider two carts, of masses m and 2m , at r | vedclass

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Question

$\Delta \mathrm{p}=\mathrm{F} \cdot \Delta \mathrm{t}$ so

momentum of lighter cart $\left(p_{1}\right)=$ momentum of heavier cart $\left(\mathrm{p}

Vedclass · Accepted Answer

larger than the kinetic energy of the heavy cart

Vedclass · Answer

$\Delta \mathrm{p}=\mathrm{F} \cdot \Delta \mathrm{t}$ so

momentum of lighter cart $\left(p_{1}\right)=$ momentum of heavier cart $\left(\mathrm{p}_{2}\right)$

$K.E._{1}$ $(lighter\, cart)$ $=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{m}}$

$K.E._{2}$ $(heavier \,cart)$ $=\frac{\mathrm{p}_{2}^{2}}{2(2 \mathrm{m})}$

So. $\mathrm{K.E}_{.1}>\mathrm{K.E.}$

Consider two carts, of masses $m$ and $2m$ , at rest on an air track. If you push both the carts for $3\,s$ exerting equal force on each, the kinetic energy of the light cart is

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