1) A truck with a mass of 1520 kg and moving with a speed of 15.0 m/s rear-ends

Question

1) A truck with a mass of 1520 kg and moving with a speed of 15.0 m/s rear-ends a 647-kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision.

2) A thin block of soft wood with a mass of 0.074 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 633 m/s at a block of wood and passes completely through it. The speed of the block is 20 m/s immediately after the bullet exits the block.
    (a) Determine the speed of the bullet as it exits the block.
    (b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy.
    (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

3) As shown in the figure below, object m1 = 1.75 kg starts at an initial height h1i = 0.260 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.75 kg which is initially at rest. Determine the following.

    (a) speed of m1 just before the collision.
    (b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.)
    (c) height to which each ball swings after the collision (ignoring air resistance)

1) A truck with a mass of 1520 kg and moving with a speed of 15.0 m/s rear-ends a 647-kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision. 2) A thin block of soft wood with a mass of 0.074 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 633 m/s at a block of wood and passes completely through it. The speed of the block is 20 m/s immediately after the bullet exits the block. (a) Determine the speed of the bullet as it exits the block. (b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy. (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system. 3) As shown in the figure below, object m1 = 1.75 kg starts at an initial height h1i = 0.260 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.75 kg which is initially at rest. Determine the following. (a) speed of m1 just before the collision. (b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.) (c) height to which each ball swings after the collision (ignoring air resistance)

Explanation / Answer

we use both conservation of momentum and energy since the collision is assumed elastic

conservation of momentum gives us:

1520kg x 14 m/s = 1520v1 + 647 v2 where v1 and v2 are the velocities fo the truck and car after collision

we can write this as

v1=(24360-647v2)/1520 (eq.1)

energy conservation gives us:

1/2(1520)(14)^2=1/2(1520)(v1)^2 +1/2(647)(v2)^2

divide through by 1/2(1520):

14^2=v1^2+0.51v2^2 or

196=v1^2+0.51v2^2 (eq.2)

substitute eq. 1 into eq. 2

v1=(24360-647v2)/1520

196=(24360-647v2)^2/1520^2 +0.43v2^2

this is a quadratic in v2; expand the right hand side and solve for v2 via the quadratic equation

this yields v2=19.6m/s and v1=5.55 m/s

both vehicles continue to move forward after the collision

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.