Question

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass 8.50 kg, is sliding to the left at 6.00 m/s, while the other, of mass 5.75 kg, is slipping to the right at 5.50 m/s. They hold fast to each other after they collide.
(a) Find the magnitude and direction of the velocity of these free-spirited otters right after they collide.
(b) How much mechanical energy dissipates during this play?

Answer 1

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value into the formula

$8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v$

$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

$K.E_{i}=239.96\ J$

We need to calculate the final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Answer 2

Answer:

Explanation:

m1 = 8.5 kg

m2 = 5.75 kg

u1 = - 6 m/s

u2 = 5.5 m/s

(a) Let the velocity after collision is v.

Use the conservation of momentum

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 8.5 x 6 + 5.75 x 5.5 = (8.5 + 5.75) x v

- 51 + 31.625 = 14.25 v

v = - 1.36 m/s

So, it moves towards left after collision.

(b) Kinetic energy before collision,

Ki = 0.5 x 8.5 x 6 x 6 + 0.5 x 5.75 x 5.5 x 5.5 = 240 J

Kinetic energy after collision

Kf = 0.5 x (8.5 + 5.75) x 1.36 x 1.36 = 13.18 J

Change in energy = 240 - 13.18 = 226.82 J