Question

A mass mAmAm_A = 1.8 kgkg , moving with velocity v⃗ A=(4.0iˆ+4.4jˆ−1.2kˆ)m/sv→A=(4.0i^+4.4j^−1.2k^)m/s, collides with mass mBmBm_B = 3.8 kgkg , which is initially at rest. Immediately after the collision, mass mAmAm_A = 1.8 kgkg is observed traveling at velocity v⃗ ′A=(−2.4iˆ+3.0kˆ)m/sv→′A=(−2.4i^+3.0k^)m/s.

Find the velocity of mass mBmB after the collision. Assume no outside force acts on the two masses during the collision.

Enter the x, y, and z components of the velocity separated by commas. Express your answer to two significant figures.

Answer 1

vB'x, vB'y, vB'z = 0.76, 2.1, 2.0 m/s <<<<<<<-----------------Answer

Let
m_A = 1.8 kg
vAx = 4 m/s
vAy = 4.4 m/s
vAz = -1.2 m/s

m_B = 3.8 kg
vBx = 0
vBy = 0
vBz = 0

After the collision,
vA'x = -2.4 m/s
vA'y = 0 m/s
vA'z = 3 m/s

let vB'x, vB'y and vB'z are the x,y and z components of B after the collision.

Apply conservation of momentum in x-direction,

final momentum = initial momentum

1.8*(-2.4) + 3.8*vB'x = 1.8*4

==> vB'x = (1.8*4 - 1.8*2.4)/3.8

= 0.76 m/s

Apply conservation of momentum in y-direction,

final momentum = initial momentum

1.8*0+ 3.8*vB'y = 1.8*4.4

==> vB'y = 1.8*4.4/3.8

= 2.1 m/s

Apply conservation of momentum in z-direction,

final momentum = initial momentum

1.8*(-1.2) + 3.8*vB'z = 1.8*3

==> vB'z = (1.8*3 + 1.8*1.2)/3.8

= 2.0 m/s

vB'x, vB'y, vB'z = 0.76, 2.1, 2.0 m/s <<<<<<<-----------------Answer