Question

Someone pls answer this

Answer 1

Answer:

The speed of metal block B is 5 m/s after the collision

Explanation:

Law Of Conservation Of Linear Momentum

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 them all

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

If some 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, we have:

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

The metal block A has a mass of m1=3.2 Kg and moves at v1=4 m/s. Metal block b has a mass of m2=1.6 Kg and is initially at rest v2=0.

After the collision occurs, block A moves at v1'=1.5 m/s. We need to calculate the speed of the metal block B. Solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

Substituting the given values:

$\displaystyle v'_2=\frac{3.2*4+1.6*0-3.2*1.5}{1.6}$

$\displaystyle v'_2=\frac{8}{1.6}$

$\displaystyle v'_2=5\ m/s$

The speed of metal block B is 5 m/s after the collision