All changes save 10. A student on a go-cart (with a combined mass of 82 kg) moving at 6.2 m collides with a 17 kg cart initially at rest. As a resul
of the collision, momentum is transferred from the first cart to the second cart. The velocity of the student and cart is 5.22"
after the collision. The collision lasts for 0.20 s. What is the absolute value of the impulse that is delivered to the student and
the first cart during the collision?
kg
Be sure to enter the absolute value of your answer. Also, round your answer to the nearest hundredth if necessary.

No because a circle or pie graph usually shows percentages you'd need a line graph

    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