Respuesta :
here we will use momentum conservation along X and Y direction.
for X direction we will have
[tex]m_1v_{1ix} + m_2v_{2ix} = m_1v_{1fx} + m_2v_{2fx}[/tex]
[tex]4*0 + 2*8 = 4*v_{1fx} + 2*4cos37[/tex]
[tex]16 - 6.4 = 4*v_{1fx}[/tex]
[tex]v_{1fx}= 2.4 m/s[/tex]
Now in Y direction we will apply momentum conservation
[tex]m_1v_{1iy} + m_2v_{2iy} = m_1v_{1fy} + m_2v_{2fy}[/tex]
[tex]4*0 + 2*0 = 4*v_{1fy} + 2*4sin37[/tex]
[tex]0 = 4*v_{1fy}+ 4.8[/tex]
[tex]v_{1fy}= -1.2 m/s[/tex]
Now net speed of the object is given by
[tex]v = \sqrt{v_{fx}^2 + v_{fy}^2}[/tex]
[tex]v = \sqrt{1.2^2 + 2.4^2}[/tex]
[tex]v = 2.68 m/s[/tex]
Answer:
The speed of the 4.0-kg mass after the collision is 2.69 m/s
Explanation:
Given data:
m₁ = mass = 4 kg
v₁ = initial speed of m₁ = 0
m₂ = 2 kg
v₂ = initial speed of m₂ = 8 m/s (x-axis)
v₂ₓ = 4 m/s with 37° x-axis
Horizontal component of final velocity v₂ₓ = v₂*cos37 = 4*cos37 = 3.195 m/s
Vertical component of final velocity v₂y = v₂*sin37 = 4*sin37 = 2.407 m/s
The momentum is conserved, then, the momentum in x-axis is:
[tex]m_{1} *0+m_{2} *8=m_{1}*v_{1x} +m_{2} *3.195\\v_{1x}=\frac{m_{2} *8-m_{2} *3.195}{m_{1}} \\v_{1x}=\frac{(2*8)-(2*3.195)}{4} =2.4m/s[/tex]
The momentum is conserved, then, the momentum in y-axis is:
[tex]m_{1} *0+m_{2} *0=m_{1}*v_{1y} +m_{2} *2.407\\v_{1y}=-\frac{m_{2}*2.407 }{m_{1} } =-\frac{2*2.407}{4} =-1.204m/s[/tex]
The final speed after the collision is:
[tex]v_{1} =\sqrt{v_{1x}^{2}+v_{1y}^{2} } =\sqrt{2.4^{2}+(-1.204)^{2} } =2.69m/s[/tex]
