91Ó°ÊÓ

A head on collision of object 1 and object 2 having mass $m_1$and $m_2$which were in rest.

A net force affects the momentum of an object whose momentum is the product of mass and velocity, according to the momentum principle.

${\mathit{F}}_{\mathbf{\text{net}}}\mathbf{=}\frac{\mathbf{Δ}\mathbf{p}}{\mathbf{Δ}\mathbf{t}}$

The statement (1) is correct as:

${\stackrel{→}{p}}_{1,\text{initial}}={\stackrel{→}{p}}_{1,\text{final}}+{\stackrel{→}{p}}_{2,\text{final}}$

Collision of will follow the momentum principle. So, the momentum before collision is equal to the momentum after collision, and${\stackrel{→}{p}}_{2,\text{initial}}=0$.

The statement (2) is correct as:

$\left|{\stackrel{→}{p}}_{1,\text{final}}\right|<\left|{\stackrel{→}{p}}_{1,\text{initial}}\right|$

From the momentum principle, the amount of${\stackrel{→}{p}}_{1,\text{initial}}$is equal to${\stackrel{→}{p}}_{1,\text{final}}+{\stackrel{→}{p}}_{2,\text{final}}$

The statement (2) is correct.

(3) The statement is,$m_2>m_1$ , then$\left|\mathrm{Δ}{P}_1\right|>\left|\mathrm{Δ}{P}_2\right|$ .

Break it down into terms so we can see if the equation is true.

$\left|\mathrm{Δ}{P}_1\right|=\left|\mathrm{Δ}{P}_2\right|m_1\left(v_1-u_1\right)=m_2\left(v_2-u_2\right)m_1\left(v_1-u_1\right)=m_2\left(v_2\right)$

If$m_2>m_1$, then$\left|\mathrm{Δ}{P}_1\right|<âˆ\pounds \mathrm{Δ}{P}_2$, which means the statement is not true.

(4) Let us check if the final speed of object 2 is less than the initial speed of the object

$\left|\mathrm{Δ}{P}_1\right|=\left|\mathrm{Δ}{P}_2\right|m_1\left(v_1-u_1\right)=m_2\left(v_2-u_2\right)m_1\left(v_1-u_1\right)=m_2\left(v_2\right)v_2=\frac{m_1\left(v_1-u_1\right)}{m_2}$

Here, if$m_1>m_2$, the final speed of the second object is greater than the first object's initial speed. This will make the statement, not true.

(5) Let us check if the final speed of object 1 is greater than the final speed of the object 2 .

$\left|\mathrm{Δ}{P}_1\right|=\left|\mathrm{Δ}{P}_2\right|m_1\left(v_1-u_1\right)=m_2\left(v_2-u_2\right)m_1\left(v_1-u_1\right)=m_2\left(v_2\right)v_2=\frac{m_1\left(v_1-u_1\right)}{m_2}$

Here, if$m_2>m_1$, the final speed of the first object is greater than the second object's final speed. This will make the statement, true.