Question

Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌ ‌It‌ ‌collides‌ ‌with‌ ‌glider‌ ‌B‌ ‌of‌ ‌mass‌ ‌0.710‌ ‌kg‌ ‌moving‌ ‌in‌ ‌the‌ ‌same‌ ‌direction‌ ‌at‌ ‌a‌ ‌speed‌ ‌of‌ ‌0.045‌ ‌m/s.‌ ‌After‌ ‌the‌ ‌collision,‌ ‌glider‌ ‌A‌ ‌continues‌ ‌in‌ ‌the‌ ‌same‌ ‌direction‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.035‌ ‌m/s.‌ ‌What‌ ‌is‌ ‌the‌ ‌velocity‌ ‌of‌ ‌glider‌ ‌B‌ ‌after‌ ‌the‌ ‌collision?

Answer 1

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

$(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})$

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

$(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]$