Question

A 30kg hunting dog leaps from a 25kg canoe where its master, a 75kg man, is sitting. The canoe is initially at rest. The dog leaps with a velocity of 4m/s. What is the velocity of the canoe just after the dog jumps?

Answer 1

Given :

Mass of the dog = 30Kg

Mass of the man = 75Kg

Velocity of the dog = 4m/s

To Find :

The velocity of the canoe after the dog jumps = ?

Solution :

• The initial momentum of the canoe is zero as both the dog and its master are in rest position

• Let us assume the canoe moves in opposite direction with a velocity ‘v’

• By principle of conservation of momentum

              Initial momentum = Final momentum

• By applying the values to the formula

         $0=m_{dog} v_{dog} +m_{man} v_{man}$

         $30\times 4+v_{man} \times 75=0$

         $v_{man}\times 75=-120$

         $v_{man} =-0.625m/s$

• The velocity of the canoe is 0.625m/s in the opposite direction.

Answer 2

Given:

A 30 kg hunting dog leaps from a 25 kg canoe where its master, a 75 kg man, is sitting. The canoe is initially at rest. The dog leaps with a velocity of 4 m/s.

To Find:

What is the velocity of the canoe just after the dog jumps?

Solution:

Mass of dog = m = 30 kg

Mass of canoe with master = M = 25 + 75 = 100 kg

$\mathbf{\textrm{Final velocity of dog}= u = 4 \ \ \dfrac{m}{s}}$

$\mathbf{\textrm{Final velocity of canoe with master}= V = Unknown \ \ \dfrac{m}{s}}$

Initial momentum of total system  = 0

Here net force acting between dog and canoe with master is zero. So it's momentum is conserved:

From conservation of momentum:

Final momentum = initial momentum

$\mathbf{M\times V+m\times u=0}$

$\mathbf{100\times V+30\times 4=0}$

On simplify:

$\mathbf{V =-1.2 \ \ \dfrac{m}{s} }$

Means velocity of canoe just after dog jump is 1.2 m/s in opposite direction of dog velocity.