Question

Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed v1 = 17.0 m/s . It collides with block 2, of mass m2 = 51.0 kg , which was initially at rest. The blocks stick together after the collision. Part A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.Part B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.Part C. What is the change of K=Kfinal-Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. if you can show your work just so I can follow along, and correct my mistakes

Answer 1

Answer:

$p=45.9\ kg.m/s$

$v_f=0.85\ m/s$

$\Delta K=-370.75 \ Joule$

Explanation:

Linear Momentum

An object of mass m traveling at a speed v in a linear path has a momentum of

$p=m.v$

It this object collides with another object and no external forces are acting on the system, then the total momentum is conserved, thus, being p the initial momentum

$p=m_1v_1+m_2v_2$

and p' the final momentum

$p'=m_1v_1'+m_2v_2'$

They must be equal, thus

$m_1v_1'+m_2v_2'=m_1v_1+m_2v_2$

Part A. The total momentum is now computed:

We have that $m_1=2.70 kg, m_2=51 kg, v_1=17 m/s, v_2=0.$

$p=(2.7)(17)+51(0)=45.9\ kg.m/s$

$\boxed{p=45.9\ kg.m/s}$

Part B.

Both blocks collide and stick together in an unique block of mass m_1+m_2:

$(m_1+m_2)v'=m_1v_1+m_2v_2$

v' is the common speed of the blocks after the collision. We now solve for v'

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

$\displaystyle v'=\frac{(2.7)(17)+0}{2.7+51}$

$\boxed{v'=0.85\ m/s}$

Part C:

The total kinetic energy before the collision is

$\displaystyle K=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$

$\displaystyle K=\frac{(2.7)(17)^2}{2}+\frac{(51)0^2}{2}$

$K=390.15\ Joule$

The total kinetic energy after the collision is

$\displaystyle K'=\frac{m_1v_1'^2}{2}+\frac{m_2v_2'^2}{2}$

$\displaystyle K'=\frac{(m_1+m_2)v'^2}{2}$

$\displaystyle K'=\frac{(2.7+51)(0.85)^2}{2}$

$K'=19.4\ Joule$

There is a change of

$\boxed{\Delta K=19.4-390.15=-370.75 \ Joule}$