a 3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterward, the 3.00 kg block moves 1.11 m/s lef

Question

3 years ago

6

t. find the momentum of the 2.22 kg block afterwards

1 answer:

LuckyWell [14K] · Answer 1 · 2022-05-27T18:21:09+03:00

The momentum of the 2.22 kg afterwards is 0.91 kg m/s to the right

Explanation:

We can solve this problem by applying the law of conservation of momentum.

In fact, the total momentum before and after the collision must be conserved. So, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 3.00 kg$ is the mass of the first block

$u_1 = 2.09 m/s$ is the initial velocity of the first block (we take the right as positive direction)

$v_1 = -1.11 m/s$ is the final velocity of the first block (to the left)

$m_2 = 2.22 kg$ is the mass of the second block

$u_2 = -3.92$ is the initial velocity of the second block

$v_2$ is the final velocity of the second block

Re-arranging the equation and substituting the values, we find: the final velocity of the second block:

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(3.00)(2.09)+(2.22)(-3.92)-(3.00)(-1.11)}{2.22}=0.41 m/s$

(to the right, since it is positive)

And so, the momentum of the 2.22 kg block afterwards is:

$p_2 = m_2 v_2 = (2.22)(0.41)=0.91 kg m/s$ (to the right)

Learn more about momentum here:

#LearnwithBrainly