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Chapter 6: Problem 3

A \(4.88\) -kg object with a speed of \(31.4 \mathrm{~m} / \mathrm{s}\) strikes a steel plate at an angle of \(42.0^{\circ}\) and rebounds at the same speed and angle (Fig. 6-19). What is the change (magnitude and direction) of the linear momentum of the object?

Short Answer

Expert verified
The change in the linear momentum of the object is \(306.624 kg·m/s\) in the opposite direction of the initial momentum.

Step by step solution

01

Calculate Initial Momentum

Calculate the initial momentum of the object using the formula \( \text{momentum} = mass \times velocity\). The initial momentum is \(4.88 kg \times 31.4 m/s = 153.312 kg·m/s\) in the direction it was originally moving.
02

Calculate Final Momentum

Calculate the final momentum of the object after it rebounds. It's still \(4.88 kg \times 31.4 m/s = 153.312 kg·m/s\), but the direction has changed i.e., it's still the same magnitude, but now in the opposite direction.
03

Calculate Change in Momentum

The change in momentum is the final momentum minus the initial momentum. Keeping in mind that the direction has changed, the change is \(153.312 kg·m/s - (-153.312 kg·m/s) = 306.624 kg·m/s\).

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Most popular questions from this chapter

Two parts of a spacecraft are separated by detonating the explosive bolts that hold them together. The masses of the parts are \(1200 \mathrm{~kg}\) and \(1800 \mathrm{~kg} ;\) the magnitude of the impulse delivered to each part is \(300 \mathrm{~N} \cdot \mathrm{s}\). What is the relative speed of separation of the two parts?

A golfer hits a golf ball, imparting to it an initial velocity of magnitude \(52.2 \mathrm{~m} / \mathrm{s}\) directed \(30^{\circ}\) above the horizontal. Assuming that the mass of the ball is \(46.0 \mathrm{~g}\) and the club and ball are in contact for \(1.20 \mathrm{~ms}\), find \((a)\) the impulse imparted to the ball, \((b)\) the impulse imparted to the club, and \((c)\) the average force exerted on the ball by the club.

Two objects, \(A\) and \(B\), collide. A has mass \(2.0 \mathrm{~kg}\), and \(B\) has mass \(3.0 \mathrm{~kg}\). The velocities before the collision are \(\overrightarrow{\mathbf{v}}_{\mathrm{iA}}=\) \((15 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{v}}_{\mathrm{i} B}=(-10 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\) After the collision, \(\overrightarrow{\mathbf{v}}_{\mathrm{fA}}=(-6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\). What is the final velocity of \(B\) ?

A \(150-\mathrm{g}\) (weight \(=5.30 \mathrm{oz}\) ) baseball pitched at a speed of \(41.6 \mathrm{~m} / \mathrm{s}(=136 \mathrm{ft} / \mathrm{s})\) is hit straight back to the pitcher at a speed of \(61.5 \mathrm{~m} / \mathrm{s}(=202 \mathrm{ft} / \mathrm{s}) .\) The bat is in contact with the ball for \(4.70 \mathrm{~ms}\). Find the average force exerted by the bat on the ball.

A ball of mass \(m\) and speed \(v\) strikes a wall perpendicularly and rebounds with undiminished speed. ( \(a\) ) If the time of collision is \(\Delta t\), what is the average force exerted by the ball on the wall? (b) Evaluate this average force numerically for a rubber ball with mass \(140 \mathrm{~g}\) moving at \(7.8 \mathrm{~m} / \mathrm{s}\); the duration of the collision is \(3.9 \mathrm{~ms}\).
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