Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.510 m/s. The total mass of the sled, man, and rock is 97.5 kg. The mass of the rock is 0.320 kg and the man can throw it with a speed of 16.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving). m/s Determine the speed of the sled (in m/s) if the man throws the rock directly backward.

Answer 1

a) v =
`v = 4.70* 10^(-3) m/s`

b)V = 0.059 m/s

Step-by-step explanation:

given data:

speed of frozen lake is 0.510 m/s

mass of rock + man is 97.5 kg

mass of rock is 0.320 kg

speed of throwing rock is 16.5 kg

By conservation of momentum...

(m + M)(v) = mv + Mv

(97.5)(0.510) = (0.320)(16.5) + (97.5 - 0.320)(v)

`v = 4.70* 10^(-3) m/s`

for backward..

(m + M)(v) = mv + Mv

`(97.5)(4.70* 10^(-3)) = -(.32)(16.5) + (97.5 - .32)(v)`

V = 0.059 m/s

Answer 2

0.45734 m/s

0.56601 m/s

Step-by-step explanation:

M+m = Mass of sled, man, and rock = 97.5 kg

m = Mass of rock = 0.320 kg

M = Mass of Man and sled = 97.5 - 0.320 = 97.18 kg

`u_1` = Velocity of rock = 16.5 m/s

`u_2` = Velocity of sled

Here the momentum of the system is conserved

`(m+M)v=mu_1+Mu_2\\\Rightarrow u_2=((m+M)v-mu_1)/(M)\\\Rightarrow u_2=((97.5)0.510-0.320* 16.5)/(97.18)\\\Rightarrow u_2=0.45734\ m/s`

The speed of the sled is 0.45734 m/s

In the case of opposite direction the speed will become negative

`(m+M)v=mu_1+Mu_2\\\Rightarrow u_2=((m+M)v-mu_1)/(M)\\\Rightarrow u_2=((97.5)0.510-(-0.320* 16.5))/(97.18)\\\Rightarrow u_2=0.56601\ m/s`

The speed of the sled is 0.56601 m/s