Question

A 4.0-kg mass, initially at rest on a horizontal frictionless surface, is struck by a 2.0-kg mass moving along the x axis with a speed of 8.0 m/s. after the collision, the 2.0-kg mass has a speed of 4.0 m/s at an angle of 37° from the positive x axis. what is the speed of the 4.0-kg mass after the collision?

Answer 1

The speed of the 4.0-kg mass after the collision is 2.69 m/s

Step-by-step 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:

`m_(1) *0+m_(2) *8=m_(1)*v_(1x) +m_(2) *3.195\\v_(1x)=(m_(2) *8-m_(2) *3.195)/(m_(1)) \\v_(1x)=((2*8)-(2*3.195))/(4) =2.4m/s`

The momentum is conserved, then, the momentum in y-axis is:

`m_(1) *0+m_(2) *0=m_(1)*v_(1y) +m_(2) *2.407\\v_(1y)=-(m_(2)*2.407 )/(m_(1) ) =-(2*2.407)/(4) =-1.204m/s`

The final speed after the collision is:

`v_(1) =\sqrt{v_(1x)^(2)+v_(1y)^(2) } =\sqrt{2.4^(2)+(-1.204)^(2) } =2.69m/s`

Answer 2

here we will use momentum conservation along X and Y direction.

for X direction we will have

`m_1v_(1ix) + m_2v_(2ix) = m_1v_(1fx) + m_2v_(2fx)`

`4*0 + 2*8 = 4*v_(1fx) + 2*4cos37`

`16 - 6.4 = 4*v_(1fx)`

`v_(1fx)= 2.4 m/s`

Now in Y direction we will apply momentum conservation

`m_1v_(1iy) + m_2v_(2iy) = m_1v_(1fy) + m_2v_(2fy)`

`4*0 + 2*0 = 4*v_(1fy) + 2*4sin37`

`0 = 4*v_(1fy)+ 4.8`

`v_(1fy)= -1.2 m/s`

Now net speed of the object is given by

`v = \sqrt{v_(fx)^2 + v_(fy)^2}`

`v = √(1.2^2 + 2.4^2)`

`v = 2.68 m/s`