Question

A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.20 kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down.
how high above the valley floor will the combined chunks go?

Answer 1

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m