91影视

The given data is listed below as,

• The mass of the projectile is, $m_1$.
• The initial velocity of the projectile is, .
• The direction in which the projectile is moving is, direction.
• The mass of the stationary target that the projectile strikes is, .
• The initial velocity of the target is, .

The law of conservation of momentum states that the momentum of a system before and after collision with another system remains constant if no external force acts on it.

Let the initial velocity of the projectile be $v_{1i}$and the initial velocity of the target is $v_{2i}$that is 0 as it was at rest and the final velocity of the projectile is $v_{1f}$and the final velocity of the target is $v_{2f}$.

From the law of conservation of momentum, the equation of the momentum of the projectile and the target can be expressed as,

role="math" localid="1657858281462" $(m_1-v_{1i})+(m_2-v_{2i})=(m_1-v_{1f})+(m_2-v_{2f})$

$\mathrm{For}{v}_{2i}=0$

$$\begin{gathered} (m_1-v_{1i})+(m_2-0)=(m_1-v_{1f})+(m_2-v_{2f}) \\ (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{2f}) \end{gathered}$$...........(1)

The expression for the velocity of the projectile and the target is as follows,

$v_{1i}+v_{1f}=v_{2i}+v_{2f}$

For $v_{2i}=0$,

$v_{1i}+v_{1f}=v_{2f}$ 鈥�(2)

For $v_{1j}+v_{1f}=v_{2f}$in the equation (1),

$$\begin{gathered} (m_1-v_{1i})=(m_1-v_{1f})+(m_2-(v_{1i}+v_{1f}\left)\right) \\ (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{1i}+m_2-v_{1f}) \end{gathered}$$

Hence, from the above equation, the value of the final velocity of the projectile becomes,

role="math" localid="1657859475686" $$\begin{gathered} (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{1i}+m_2-v_{1f}) \\ {v}_{1f}(m_1+m_1)=m_1{v}_{1f}-m_2{v}_{1f} \\ {v}_{1f}=\frac{v_{1f}(m_1-m_2)}{m_1+m_2} \end{gathered}$$

From the equation (2), the value of the final velocity of the projectile can be expressed as,

$v_{1f}=v_{2f}-v_{1i}$

For $v_{1f}=v_{2f}-v_{1i}$in the equation (1),

$$\begin{gathered} (m_1-v_{1i})=(m_1.(v_{2f}-v_{1f}\left)\right)+(m_2-v_{2f}) \\ (m_1-m_1)=m_1{v}_{2f}-m_2{v}_{1i}+m_2{v}_{2f} \\ {v}_{1f}\left(2m_1\right)=v_{2f}(m_1+m_2) \\ {v}_{1f}=\frac{2m{v}_{1f}}{(m_1+m_2)} \end{gathered}$$

From the gathered value of the final velocity of the projectile, it is evident that the value of $m_1$is larger than the value of$m_2$, otherwise the velocity will be negative which cannot be possible.

Thus, the value of the final velocity of the projectile is $\frac{v_{1j}(m_1-m_2)}{m_1+m_2}$, the final velocity of the target is $\frac{2m_1{v}_{1j}}{m_1+m_2}$, and the final velocity of the projectile shows that $m_1鈮�m_2$.