Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

# Help!I don't understand these physics problems.?

2.A 77.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 19.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)

a.puck ____ m/s

b.goalie ____ m/s

2.In an ice show a 45.0 kg skater leaps into the air and is caught by an initially stationary 55.0 kg skater.

(a) What is their final velocity assuming negligible friction and that the leaper's original horizontal velocity was 4.00 m/s?

__________ m/s

(b) How much kinetic energy is lost?

______J

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• 2 months ago

You labeled both questions as #2.  I'm going to do the last question first:

(45 kg)(4.00 m/s) = (100 kg)(final velocity) =>

final velocity = 1.80 m/s.

(b) The loss of kinetic energy is

(1/2)(45 kg)(4.00 m/s)^2 - (1/2)(100 kg)(1.80 m/s)^2

= 360 J - 162 J = 198 J.

Now the "puck and goalie" question.  I'll call their final velocities vp and vg.

(0.150 kg)(19.0 m/s) = (0.150 kg)vp + (77.0 kg)vg and also

(1/2)(0.150 kg)(19.0 m/s)^2 =

= (1/2)(0.150 kg)(vp)^2 + (1/2)(77.0 kg)(vg)^2.

From the first equation you have

(0.150)(19.0 m/s - vp) = (77.0 kg)(vg) =>

(vg) = 0.03701 m/s - 0.001948 vp.

Substitute that expression for the "vg" in the 2nd (quadratic) equation, and solve the equation for vp, which will be the only unknown.  There will be two solutions; use the negative value of vp.

After you have vp, then vg = 0.03701 m/s - 0.001948 vp.