Chapter 11: Impulse and Momentum-Part 2

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is

1 m/s.

You are given the equation used to solve a problem: (0.10kg)(40m/s)−(0.10kg)(−30m/s)=12(1400N)Δt.

A 0.10 kg ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in contact with the bat? B. 1.0×10^-2 s

You are given the equation used to solve a problem: (600g)(4.8m/s)=(400g)(3.0m/s)+(200g)(vix)2.

A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to the right at 3.0 m/s. The balls stick and move to the right at 4.8 m/s. What was the speed of the 200 g ball of clay? B. 8.4 m/s

A force pushes the cart for 1 s, starting from rest. To achieve the same speed with a force half as big, the force would need to push for

2s

In a collision between two objects having unequal masses, how does magnitude of the impulse imparted to the lighter object by the heavier one compare with the magnitude of the impulse imparted to the heavier object by the lighter one?

Both objects receive the same impulse.

A 20 g ball of clay traveling east at 2.0m/s collides with a 30 g ball of clay traveling 30∘∘ south of west at 1.0m/s. What is the speed of the resulting 50 g blob of clay?

Masses m = 20 g M = 30 g Initial velocities u = 2i U = 1x cos 30 (-i) + 1 x sin 30 ( -j) = -0.866 i -0.5 j 40 i -25.98 i - 15 j = 50 v 14.01 i - 15 j = 50 v v = 0.2803 i - 0.3 j Speed of the resulting 50 g clay v = √[ 0.2803 2 + (-0.3) 2 ] = 0.4105 m / s (b). Let v makes an angle θ with east along south then tan θ = 0.3 / 0.2803 θ = 46.94 o south of east

In an inelastic collision, which of the following quantities is conserved?

Momentum

A 20.0 kg wood ball hangs from a 1.50 m-long wire. The maximum tension the wire can withstand without breaking is 600 N. A 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball.

The given parameters; mass of the wood, m = 20 kg length of the wire, l = 1.3 m maximum tension on the wire, T = 600 N mass of the projectile, = 0.9 kg Apply the principle of conservation of linear momentum; 20(0)+0.9 u2 = v(20+0.9) u2= 20.9/0.9 u2 = 23.2v The net force of the wood-projectile system in the circular path is calculated as follows; 600-9.8(20+0.9) = (20+0.9) x v^2/1.5 395.18 = 13.93v^2 v^2=28.37 v= sqrt 28.37 v=5.326 The largest speed the projectile can have: u2 = 23.2 x 5.326 u2 = 124 m/s

The figure(Figure 1) shows a collision between three balls of clay. The three hit simultaneously and stick together.

What is the speed of the resulting blob of clay? v=0.85 m/s What is the direction of the resulting blob of clay? 72 degree

Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v1, and car 2 was traveling northward at a speed of v2. (Figure 1)After the collision, the two cars stick together and travel off in the direction shown.

a. [sqrt(m2v2)^2 + (m1v1)^2]/(m1+m2) b. tan = m2v2/m1v1 c. The magnitudes of the momenta of the cars were equal.

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle ϕ south of east (as indicated in the figure). After the collision, the two-car system travels at speed vfinal at an angle θ east of north

a. vfinal = sqrt 5v^2/4 - v^2 sinϕ b. tan^-1(cosϕ / 2-sinϕ)

A proton (mass 1 u) is shot toward an unknown target nucleus at a speed of 2.40×106 m/s . The proton rebounds with its speed reduced by 25% while the target nucleus acquires a speed of 3.50×105 m/s .

m = 12 u

A perfectly elastic collision conserves ________.

mass mechanical energy momentum

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring between them. They are not attached to the spring. After they are released and have both moved free of the spring

the lighter block will have more kinetic energy than the heavier block.

Force Fx=(10N)sin(2πt/4.0s) is exerted on a 370 g particle during the interval 0s≤t≤2.0s.

If the particle starts from rest, what is its speed at t=2.0s? 1/0.370 x 2/pi x 10 x 2 v = 34 m/s

When is the total momentum of a system conserved?

If the system is isolated

A perfectly elastic collision is a collision:

That conserves mechanical energy.

A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct?

The small car and the truck experience the same average force.

In a perfectly ELASTIC collision between two perfectly rigid objects

both the momentum and the kinetic energy of the system are conserved.

Impulse is

the area under the force curve in a force-versus-time graph.

In an INELASTIC collision between two objects

the momentum of the system is conserved but the kinetic energy of the system is not conserved.

Jacques and George meet in the middle of a lake while paddling in their canoes. They come to a complete stop and talk for a while. When they are ready to leave, Jacques pushes George's canoe with a force F⃗ �→ to separate the two canoes. What is correct to say about the final momentum and kinetic energy of the system(consisting of the two canoes and the men in them) if we can neglect any resistance due to the water?

The final momentum is zero but the final kinetic energy is positive.