Question

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 18.5 kg and an initial velocity of v0A = 8.15 m/s, due east. Object B, however, has a mass of mB = 30.5 kg and an initial velocity of v0B = 5.00 m/s, due north. Find the magnitude of the final velocity of the two-object system after the collision.

Answer 1

Answer:

v =4.36 m/s

Explanation:

given,

mass of object A = 18.5 Kg

initial velocity of object A = 8.15 m/s in east

mass of object B = 30.5 kg

initial velocity of object B = 5 m/s

$P = P_A+P_B$

$P = m_Av_A\widehat{i} + m_B v_B\widehat{j}$

$P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}$

$P = 150.775 \widehat{i} + 152.5 \widehat{j}$

$P = \sqrt{150.775^2+152.5^2}$

P = 214. 45 N s

velocity after collision is equal to

$v =\dfrac{214.45}{18.5+30.5}$

v =4.36 m/s

hence, velocity after collision is equal to 4.36 m/s

Answer 2

Answer:

The magnitude of the final velocity of the two-object system is $v=4.37\frac{m}{s}$

Explanation:

As the Momentum is conserved, we can compare the instant before the collision, and the instant after. Also, we have to take in account the two components of the problem (x-direction and y-direction).

To do that, we put our 0 of coordinates where the collision takes place.

So, for the initial momentum we have that

$p_{ix}=m_{a}v_{0a}+0$

$p_{iy}=0+m_{b}v_{0b}$

Now, this is equal to the final momentum (in each coordinate)

$p_{fx}=(m_{a}+m_{b}) v_{fx}$

$p_{fy}=(m_{a}+m_{b}) v_{fy}$

So, we equalize each coordinate and get each final velocity

$m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}$

$m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}$

Finally, to calculate the magnitude of the final velocity, we need to calculate

$v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}$

which, replacing with the previous results, is

$v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}=(\sqrt{(\frac{18.5*8.15}{49})^{2}+(\frac{30.5*5.00}{49})^{2}})\frac{m}{s}$

Therefore, the outcome is

$v_{f}=4.37\frac{m}{s}$