High-speed stroboscopic photographs show that the head of a 230 g golf club is t

Question

High-speed stroboscopic photographs show that the head of a 230 g golf club is traveling at 43 m/s just before it strikes a 46 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 31 m/s. Find the speed of the golf ball just after impact. m/s A 775 N man stands in the middle of a frozen pond of radius 10.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2 kg physics textbook horizontally toward the north shore at a speed of 10.0 m/s. How long does >>t take him to reach the south shore? s

Explanation / Answer

given:
m = 230 g = 0.230 kg
v = 43 m/s
mass of golf ball =46 g = 0.046 kg
velocity = 31 m/s
law of conservation of momentum
mass1 (FinalVelocity1 - InitialVelocity1)= - [mass2 (FinalVelocity2 - InitialVelocity2) ]

0.230* ( 31 - 43) = 0.046 (Vf2 - 0)
Vf2 = -60 m/s

2. given:

F =775N
radius = 10 m
mass =1.2 kg
speed =10 m/s
conservation of linear momentum says that "total momentum of a closed system of objects (which has no interactions with external agents) is constant", so:

MV = mv
where:
M - mass of the man
V - velocity of the man
m - mass of the textbook
v - velocity of the textbook

If 775 N means force that the man acts on the frozen ponds (it's a little strange for me but ...), his mass is M=775/9.81 kg (9.81 m/(s*s) g - gravitational acceleration) = 79 kg

V = mv/M = 1.2*10/79 = 0.1518 m/s

t= s/v (s - distance)

t= 10/0.1518 = 65 s

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