Two perpendicular displacements: 7 m south and 24 m east. Enter the resultant's magnitude in m.
A 1000 kg car slides east through an icy intersection at 9 m/s as a 1500 kg car slides north at 8 m/s. They hit and lock together. The wreck moves off between east and north. This lesson puts a number on its speed.
Each direction keeps its own books, so the pair's momentum has an eastward part and a northward part. Those two parts are perpendicular, and the Pythagorean theorem turns two perpendicular parts into one total size.
Take east and north as the positive directions. The eastbound car's momentum is 1000 × 9 = 9000 kg·m/s, all of it eastward. The northbound car's is 1500 × 8 = 12000 kg·m/s, all of it northward.
After the cars lock, the pair still carries 9000 kg·m/s eastward and 12000 kg·m/s northward.
The two parts are perpendicular: they form the two legs of a right triangle whose longest side is the pair's total momentum.
So the total's size comes from the Pythagorean theorem: the square root of 9000² + 12000², which is 15000 kg·m/s.
The locked pair's mass is 1000 + 1500 = 2500 kg. Velocity is momentum divided by mass: v' = 15000 ÷ 2500 = 6, so the wreck slides off at 6 m/s.
This routine covers exactly one case: two objects at a right angle that stick together. Angled, non-perpendicular collisions are beyond this course.
Take east and north as the positive directions. A 3 kg cart moving east at 2 m/s and a 2 kg cart moving north at 4 m/s collide at a right angle and couple. Calculate the pair's speed after the collision.
Write down the values in the question:
eastward momentum = 3 × 2 = 6 kg·m/s
northward momentum = 2 × 4 = 8 kg·m/s
combined mass = 3 + 2 = 5 kg
Write down the equation:
p_total = square root of (p_east² + p_north²), then v' = p_total ÷ (m_1 + m_2)
Substitute in the values:
p_total = square root of (6² + 8²) = square root of (36 + 64) = square root of 100
p_total = 10 kg·m/s
Simplify:
v' = 10 ÷ 5 = 2
v' = 2 m/s
The coupled pair moves off at 2 m/s, between east and north.
Take east and north as the positive directions. A 0.5 kg puck moving east at 6 m/s and a 0.5 kg puck moving north at 8 m/s collide at a right angle and stick. Calculate the pair's speed after the collision.
Write down the values in the question:
eastward momentum = 0.5 × 6 = 3 kg·m/s
northward momentum = 0.5 × 8 = 4 kg·m/s
combined mass = 0.5 + 0.5 = 1 kg
Write down the equation:
p_total = square root of (p_east² + p_north²), then v' = p_total ÷ (m_1 + m_2)
Substitute in the values:
p_total = square root of (3² + 4²) = square root of (9 + 16) = square root of 25
p_total = 5 kg·m/s
Simplify:
v' = 5 ÷ 1 = 5
v' = 5 m/s
The stuck pair moves off at 5 m/s, between east and north.
Take east and north as the positive directions. A 4 kg cart moving east at 3 m/s and a 1 kg cart moving north at 5 m/s collide at a right angle and couple. Enter the size of the pair's total momentum in kg·m/s.
Take east and north as the positive directions. A 6 kg cart moving east at 2 m/s and a 2 kg cart moving north at 8 m/s collide at a right angle and couple. Enter the pair's speed after the collision in m/s.
Margit says: 'For a right-angle sticking collision, the total momentum's size can be found with the Pythagorean theorem, because the eastward and northward totals are perpendicular.' Is she correct?
Take east and north as the positive directions. A 2 kg cart moving east at 4.5 m/s and a 2 kg cart moving north at 6 m/s collide at a right angle and couple. Enter the pair's speed after the collision in m/s.
Take east and north as the positive directions. A 4 kg dog sled model moving east at 2 m/s and a 5 kg model moving north at 3 m/s collide at a right angle and latch together. Enter the size of the pair's total momentum in kg·m/s.
In the right-angle sticking collision routine, why may the eastward and northward momenta be combined with the Pythagorean theorem?