Question

A massive truck of 1200N moving with a velocity of 2m/s hits a stationary mass of 30N. if the both bodies move together after the collision, determine their common velocity.​

Answer 1

Answer:

The common speed is 1.95 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of all of them:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses, the law of conservation of linear momentum

is written as:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

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

The common velocity after this situation is:

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

The truck of m1=1200 N (weight) travels at v1=2 m/s and hits a stationary mass (v2=0) of m2=30 N (weight). After the bodies collide, they keep moving together. Before we can calculate the common speed, we need to calculate the masses of the bodies, since they are given as weights.

$m_1=\frac{P_1}{g}=\frac{1200}{9.8}=122.45 Kg$

$m_2=\frac{P_2}{g}=\frac{30}{9.8}=3.06 Kg$

Now calculate the common speed:

$\displaystyle v'=\frac{122.45 * 2+3.06 * 0}{122.45+3.06}$

$\displaystyle v'=\frac{244.9}{125.51}=1.95\ m/s$

The common speed is 1.95 m/s