Within which type of system is the total mass conserved but not the total energy

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s

Explanation:

It is given that,

Mass of the car, $m_1=1.1\times 10^3\ kg$

Mass of the truck, $m_2=2.3\times 10^3\ kg$

Initial velocity of the car, $u_1=22\ m/s$

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2$

$v_2=15.7\ m/s$

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0

3 years ago

Why is the perfect mechine is not possible in the real life?

Vinvika [58]

Answer:

Because the mechanical advantage of the machine is affected by friction and weight but velocity ratio is not. So, mechanical advantage is less than velocity rate. Thus, the machine's efficiency is less than 100% and can't be a perfect machine

3 0

3 years ago

. When using straight ladders, the base of the ladder should be _____ away from the wall or other vertical surface for every ___

guapka [62]

one foot; four feet is the answer

3 0

3 years ago

Read 2 more answers

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. the sharp upward

ladessa [460]

<u>Answer:</u>

The skier’s launch speed = 17.08 m/s

<u>Explanation:</u>

The motion of skiing is projectile motion.

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.