Question

In a collision experiment, the ratio of the velocity change between two carts of equal inertia is found to equal 1.

Part A

What happens to this ratio if the inertia of each cart is doubled, and the same initial velocities are used?

A. The ratio of velocity changes remains one.
B. The ratio of velocity changes is halved
C. The ratio of velocity changes is doubled.

What happens to this ratio if the original value of inertia is used, and the initial velocities of the carts are doubled?

A. The ratio of velocity changes doubted
B. The ratio of velocity changes a halved.
C. The ratio of velocity changes one.

Answer 1

Answer:

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

Explanation:

A vehicle crash problem must be solved with the equation of the moment,

Initial instant Before crash

              p₀ = m v₁ + mv₂

After the crash

            $p_{f}$ = m $v_{1f}$ + m $v_{2f}$

           p₀ = $p_{f}$

If the speed ratio before and after the crash is one

           p₀ / $p_{f}$ = 1

We can assume that initially one of the cars was stopped

           m v₁₀ = m $v_{2f}$

           v₁₀ = $v_{2f}$

For the two speeds to be equal, the masses of the vehicles must be the same.

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

Answer 2

Part A. If the masses of both the cart are the same before and after collision then the ratio of velocity changes will be 1.

Part B. If the masses of both the cart are the same and their velocities are doubled before and after collision then the ratio of velocity changes will be 1.

How do you calculate the ratio of velocity changes for both the carts?

The linear momentum for the two carts is given below.

$p=m_1\times v_1 + m_2 \times v_2$

Where m1 is the mass of cart 1, m2 is the mass of cart 2. v1  and v2 are the velocity of cart 1 and cart 2 respectively.

After the collision, the linear momentum is given as,

$p' = mv'_1 +mv'_2$

Let us consider that both the cart has equal mass m. Then the linear momentum before and after the collision is given below.

$p = mv_1 +mv_2$

$p'=mv'_1 +mv'_2$

Given that the ratio of velocity for both the cart before and after the collision is 1.

$p=p'$

$\dfrac {p}{p'} =1$

Let us consider that one of two carts is traveling at one time. Then

$p=p'$

$mv_2=mv'_1$

Hence we can conclude that for the same velocity before and after the collision, both the carts will have equal mass.

Part A.

If the inertia of each cart is doubled it means its linear momentum will be twice. In this case, the ratio of velocity changes remains one as it fulfills the above condition.

$2p=2p'$

$\dfrac {2p}{2p'} =1$

Hence we can conclude that If the masses of both the cart are the same before and after collision then the ratio of velocity changes will be 1.

Part B.

If the velocity of the carts are doubled then the ratio of linear momentum is,

$\dfrac {p}{p'} =1$

In this case, if the velocities of the cart are twice then their ratio will be one if both the cart has equal mass.

Hence we can conclude that If the masses of both the cart are the same and their velocities are doubles before and after collision then the ratio of velocity changes will be 1.

To know more about the velocity, follow the link given below.

https://brainly.com/question/862972.