Question

A body of mass M moving with a speed V collides on a surface at an angle $45$ degree without changing its speed the change in momentum of the body will be?

• A. $MV(\hat{j}-\hat{i})$
• B. $MV(\hat{i}+\hat{j})$
• C. $2MV\hat{j}$
• D. $\sqrt{2}MV\hat{j}$

Answer 1

The correct option is D $\sqrt{2}MV\hat{j}$

Here, $V_i$ is the incident velocity

i.e. $V_i=vcos\left({45}^o\right)\hat{i}-vsin\left({45}^o\right)\hat{j}$

$V_r$ is the reflected velocity

i.e. $V_r=vcos\left({45}^o\right)\hat{i}+vsin\left({45}^o\right)\hat{j}$

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The momentum of the mass before collision is as follows:

$P_b=M\left[vcos\right({45}^o)\hat{i}-vsin({45}^o\left)\widehat{(}j\right)]$

$P_b=\left[\right(Mvcos\left({45}^o\right)\hat{i}-Mvsin\left({45}^o\right)\widehat{(}j\left)\right]$

The momentum of the mass after collision is as follows:

$P_a=M\left[vcos\right({45}^o)\hat{i}+vsin({45}^o\left)\widehat{(}j\right)]$

$P_a=\left[\right(Mvcos\left({45}^o\right)\hat{i}+Mvsin\left({45}^o\right)\widehat{(}j\left)\right]$

so, the change in the momentum of the mass is,

$P_a-P_b=\left[\right(Mvcos\left({45}^o\right)\hat{i}+Mvsin\left({45}^o\right)\widehat{(}j\left)\right]-\left[\right(Mvcos\left({45}^o\right)\hat{i}-Mvsin\left({45}^o\right)\widehat{(}j\left)\right]$

$P_a-P_b=2Mvsin\left({45}^o\right)\widehat{(}j)$

$P_a-P_b=\sqrt{2}Mv\widehat{(}j)$