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djyliett [7]
3 years ago
10

The total mechanical energy of a system is equally divided between kinetic energy and potential energy. can never be negative. i

s constant only if conservative forces act. is either all kinetic energy or all potential energy, at any one instant. is constant only if nonconservative forces act. A skier, of mass 60. kg, pushes off the top of a frictionless hill with an initial speed of 4.0 m/s. How fast will she be moving after dropping 10. m in elevation? 15. m/s 0.15 km/s 0.20 km/s 49. m/s 10. m/s In an inelastic collision, the final total momentum is less than the initial momentum. the same as the initial momentum. insufficient information to answer more than the initial momentum. A handball of mass 0.10 kg, traveling horizontally at 25. m/s, strikes a wall and rebounds at 19. m/s. What is the change in the momentum of the ball? 1.2 N-s 72 N-s 5.4 N-S 1.8N-s 4.4 N-s Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the momentum after the collision if car B was initially at rest? 10. KN-s to the right zero 6.0 KN-s to the right 8.0 KN-s to the left 2.0 KN-s to the right
Physics
1 answer:
inn [45]3 years ago
8 0

Answer:

1-  is constant only if conservative forces act

2- 60*9.8*10+0.5*60*4^2=0.5*60*v^2

v=14.56=15 m/s

3- B (Momentum is conserved)

4- Change in momentum= m(v-u) = 0.10(19--25)=4.4 Ns

5- 1000*6+0 = 6000 KN Momentum before to the right, So final momentum should also be same.

Explanation:

5 questions are in 1 question all of them are answered.

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2 years ago
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
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Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

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