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An object of $m \mathrm{~kg}$ with speed of $v \mathrm{~m} / \mathrm{s}$ strikes a wall at an angle $\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be
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The correct answer is:
$2 m v \cos \theta$
$\vec{P}_1=m v \sin \hat{\theta} \hat{i}-m v \cos \theta \hat{j}$
and $\vec{P}_2=m v \sin \theta \hat{i}+m v \cos \theta \hat{j}$
So change in momentum
$\overrightarrow{\Delta P}=\vec{P}_2-\vec{P}_1=2 m v \cos \theta \hat{j},|\Delta \vec{P}|=2 m v \cos \theta$
and $\vec{P}_2=m v \sin \theta \hat{i}+m v \cos \theta \hat{j}$
So change in momentum
$\overrightarrow{\Delta P}=\vec{P}_2-\vec{P}_1=2 m v \cos \theta \hat{j},|\Delta \vec{P}|=2 m v \cos \theta$
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