Question

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. it suddenly collides directly with a stationary seal of mass 200 kg. the seal grabs onto the whale and holds fast. what is the momentum of these two sea creatures just after their collision? you can neglect any drag effects of the water during the collision

Answer 1

M = mass of the whale = 1000 kg

m = mass of the seal = 200 kg

V = initial velocity of whale before collision with the seal = 6.0 m/s

v = initial velocity of the seal before collision with the whale = 0 m/s

V' = final velocity of two sea creatures after collision = ?

Using conservation of momentum

M V + m v = (M + m) V'

inserting the above values in the equation

(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'

6000 kgm/s + 0 kgm/s = (1200 kg) V'

V' = (6000 kgm/s ) /(1200 kg)

V' = 5 m/s

Answer 2

Pf = 6000 kg*m/s

Step-by-step explanation:

Using the conservation of the linear momentum:

`P_i = P_f`

Also:

`P_i=M_bV_b`

`P_f = (M_b+M_s)V_s`

Replacing:

`M_bV_b = (M_b+M_s)V_s`

where
`M_b` is the mass of the whale,
`V_b` is the velocity of te whale,
`M_s` is the mass of the seal and
`V_s` is the velocity of both after the collition.

so:

`(1000 kg)(6 m/s) = (1000 kg+ 200kg)V_s`

Solving for
`V_s`:

`V_s = 5 m/s`

Finally for find the momentum we will use the next equation:

`P_f = (M_b+M_s)V_s`

Pf = (1000+200)(5 m/s)

Pf = 6000 kg*m/s

Alternative:

we know that the linear momentum is conserved so, we only have to know the initial momentum for have the answer:

`P_i=M_bV_b`

Pi = (1000)(6 m/s)

Pi = 6000 Kg*m/s