A bullet of mass m moving with velocity v str | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the combined system:$mv =

Vedclass · Accepted Answer

$\frac{1}{2}mv^2 	imes \frac{m}{(m + M)}$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the combined system:$mv = (m + M)V$where $V$ is the final velocity of the combined system.$V = \frac{mv}{m + M}$The kinetic energy $K$ of the combined system is given by:$K = \frac{1}{2}(m + M)V^2$Substituting the value of $V$:$K = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2$$K = \frac{1}{2}(m + M) \frac{m^2v^2}{(m + M)^2}$$K = \frac{1}{2} \frac{m^2v^2}{m + M}$$K = \left( \frac{1}{2}mv^2 \right) \left( \frac{m}{m + M} \right)$

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ and gets embedded in it. The kinetic energy of the system is:

Similar Questions

$A$ body of mass $0.50 \ kg$ is moving on a smooth surface with a speed of $2.00 \ m/s$. It collides with another body of mass $1.00 \ kg$ at rest,and they move together as a single body. The energy loss during the collision is ....... $J$.

In perfectly inelastic collisions,the relative velocity of the bodies

$A$ particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2m$ and sticks to it. What is the combined velocity of the system?

$A$ bullet strikes a solid block resting on a frictionless horizontal table and gets embedded in it. Which of the following is conserved?

$A$ bag of sand of mass $9.8 \, kg$ is suspended by a rope. $A$ bullet of mass $200 \, g$ travelling with a speed of $10 \, ms^{-1}$ gets embedded in it. The loss of kinetic energy will be $... J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module