a. A 6.0 kg ball moving at 7.0 m/s strikes an 95 kg man on the chest and bounces

Question

a. A 6.0 kg ball moving at 7.0 m/s strikes an 95 kg man on the chest and bounces back at 2.0 m/s. If the man (who is wearing high quality roller blades reducing friction to a negligible amount) was originally at rest, what is his final velocity?

b. Who is developping more power, a 95 kg sprinter going from 0 m/s to 12 m/s over 45 meters, or a 110 kg person lifting a 210 kg weight 1.6 meters in 4 seconds? You must show work.

c. Two carts were intially at rest and side-by-side between two photogates. The mass of cart 1 was 0.605 kg. The spring was activated and the two carts moved away from one another, passing through the photogates and producing the reading as ....

v1 = 0.75 m/s v2= 0.34 m/s Determine the mass of cart 2.

Explanation / Answer

(a)m1= 6kg v1=7 m/s, v2f=-2 m/s

m2=95 kg,V2=0

let final velocity of man= Vm

using the law of conservation of linear momentum

m1 v1 + m2 v2= m1 V1f + m2 vm

6(7)+95(0)=6(-2)+ 95 Vm

Vm= 0.568 m/s

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