Question

A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in the toboggan. If KE1 is the original kinetic energy of the toboggan and KE2 is the kinetic energy after the parcel has been added, what is the ratio KE2/KE1.

Answer 1

Answer:

7V2^2/5V1^2

Explanation:

The original kinetic energy of the body = 1/2 mV1^2

Where m equals the mass of the toboggan and V1 = Original velocity

After the mass of 20kg parcel was drop on it, the new mass = 20kg + 50kg = 70kg

Since according to the equation V^2 is inversely proportional to mass, the speed (velocity will also change which equals V2^2

So the new kinetic energy = 1/2 m(70kg) * V2^2

The ratio of KE2/KE1= (1/2* 70kg*V2^2)/(1/2*50*V1^2)= 7V2^2/5V1^2

Answer 2

Answer:

the ratio KE2/KE1 is 0.71

Explanation:

given information:

toboggan's mass, $m_{1}$ = 50 kg

parcel's mass, $m_{2}$ = 20 kg

$m_{tot}$  = 50+20 = 70 kg

KE = $\frac{1}{2} mv^{2}$

KE2/KE1 = $\frac{1}{2} m_{1}v_{1}^{2}$/$\frac{1}{2} m_{tot}v_{2}^{2}$

              = $\frac{m_{1} }{m_{tot} }$

              = $\frac{5}{7}$

              = 0.71