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Chapter 9: Problem 79

A 14 -kg projectile is launched at \(380 \mathrm{m} / \mathrm{s}\) at a \(55^{\circ}\) angle to the horizontal. At the peak of its trajectory it collides with a second projectile moving horizontally, in the opposite direction, at \(140 \mathrm{m} / \mathrm{s} .\) The two stick together and land \(9.6 \mathrm{km}\) horizontally downrange from the first projectile's launch point. Find the mass of the second projectile.

Short Answer

Expert verified
Once you have substituted all your values into the momentum conservation equation, it's a simple arithmetic problem to solve for \(m_2\), the mass of the second projectile.

Step by step solution

01

Breakdown of the first projectile's motion

The problem can be broken down into x and y-components. The x-component of the velocity, \(V_{1x}\), can be found by using the equation \(V_{1x} = V_{1} \cos(\theta)\) where \(V_1 = 380 m/s\)and \(\theta = 55^{\circ}\). Similarly, the y-component of velocity, \(V_{1y}\), in the y-direction can be found by using the equation \(V_{1y} = V_{1} \sin(\theta)\). The total distance travelled in the x-direction, \(d\), is given in the problem \(d = 9.6 km = 9600 m\).
02

Calculate time of flight

The time of flight, \(t\), can be computed using the distance travelled in the x-direction and the x-component of the velocity. The time is given by the formula \(t = \frac{d}{V_{1x}}\).
03

Calculate the velocity of the combined projectiles

After the collision, the two projectiles move as a single object. The velocity of the combined object just before it hits the ground is equal to the x-component of the first projectile's initial velocity as there are no horizontal forces acting after the collision.
04

Use the conservation of momentum

Using the conservation of momentum which states that the total momentum before the collision is equal to the total momentum after the collision in the absence of external forces, the mass of the second projectile (\(m_2\)) can be found using the formula \(m_1V_1 + m_2V_2 = (m_1+m_2) V_f\) where \(m_1 = 14 kg\) is the mass of the first projectile, \(V_1 = 380 m/s\) is the speed of the first projectile, \(V_2 = -140 m/s\) is the speed of the second projectile and \(V_f\) is the x-component of the first projectile's initial velocity.

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Most popular questions from this chapter

In a railroad switchyard, a 56 -ton freight car is sent at \(7.0 \mathrm{mi} / \mathrm{h}\) toward a 31 -ton car moving in the same direction at \(2.6 \mathrm{mi} / \mathrm{h} .\) (a) What's the speed of the cars after they couple? (b) What fraction of the initial kinetic energy was lost in the collision?

An 11,000 -kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant \(k=0.32 \mathrm{MN} / \mathrm{m}\). The car is hit by a second car of \(9400-\mathrm{kg}\) mass moving at \(8.5 \mathrm{m} / \mathrm{s}\), and the two couple together. Find (a) the maximum compression of the spring and (b) the speed of the two cars when they rebound together from the spring.

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Two objects moving in opposite directions with the same speed \(v\) undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses.
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