Respuesta :
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
[tex]P = P_A+P_B[/tex]
[tex]P = m_Av_A\widehat{i} + m_B v_B\widehat{j}[/tex]
[tex]P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}[/tex]
[tex]P = 150.775 \widehat{i} + 152.5 \widehat{j}[/tex]
[tex]P = \sqrt{150.775^2+152.5^2}[/tex]
P = 214. 45 N s
velocity after collision is equal to
[tex]v =\dfrac{214.45}{18.5+30.5}[/tex]
v =4.36 m/s
hence, velocity after collision is equal to 4.36 m/s
Answer:
The magnitude of the final velocity of the two-object system is [tex]v=4.37\frac{m}{s}[/tex]
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
[tex]p_{ix}=m_{a}v_{0a}+0[/tex]
[tex]p_{iy}=0+m_{b}v_{0b}[/tex]
Now, this is equal to the final momentum (in each coordinate)
[tex]p_{fx}=(m_{a}+m_{b}) v_{fx}[/tex]
[tex]p_{fy}=(m_{a}+m_{b}) v_{fy}[/tex]
So, we equalize each coordinate and get each final velocity
[tex]m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}[/tex]
[tex]m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}[/tex]
Finally, to calculate the magnitude of the final velocity, we need to calculate
[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}[/tex]
which, replacing with the previous results, is
[tex]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}[/tex]
Therefore, the outcome is
[tex]v_{f}=4.37\frac{m}{s}[/tex]