Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time, when they're not sleeping or eating, joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 0.0157 kg, 0.0151 kg, 0.0233 kg, and 0.0207 kg. At this very instant, the x‑ and y‑ components of the mice's velocities are, respectively, (0.787 m/s,−0.617 m/s) , (−0.571 m/s,−0.137 m/s) , (0.577 m/s,0.205 m/s) , and (−0.839 m/s,0.561 m/s) . Calculate the x‑ and y‑ components of Bryce's mice's total momentum, px and py .

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Because there is no air in the jar, the vibrations can not travel through the air and create sound.  (that's why there is no sound in space, there is no air)