A. Anobject with mass m and speed v relative to an observer explodesinto two pie

Question

A. Anobject with mass m and speed v relative to an observer explodesinto two pieces. One piece is three times as massive as the other.The less massive piece stops relative to the observer.
How much kinetic energy is added to the system during theexplosion, with respect to the observer’sreference frame. ---------------------------------------------------------------------------------------------------------
B. A boy throws a ball with a speed of 12m/sat a target that is height h=5.00m above the level at which hereleases the ball. He wants the ball's velocity to be horizontal atthe instant it reaches the target. i. At what angle q above the horizontal must he throw theball?
ii. What is the horizontal distance from the release point to thetarget? iii. What is the speed of the ball just as it reached thetarget? --------------------------------------------------------------------------------------------------------- A. Anobject with mass m and speed v relative to an observer explodesinto two pieces. One piece is three times as massive as the other.The less massive piece stops relative to the observer.
How much kinetic energy is added to the system during theexplosion, with respect to the observer’sreference frame. ---------------------------------------------------------------------------------------------------------
B. A boy throws a ball with a speed of 12m/sat a target that is height h=5.00m above the level at which hereleases the ball. He wants the ball's velocity to be horizontal atthe instant it reaches the target. i. At what angle q above the horizontal must he throw theball?
ii. What is the horizontal distance from the release point to thetarget? iii. What is the speed of the ball just as it reached thetarget? ---------------------------------------------------------------------------------------------------------

Explanation / Answer

1)The mass and speed of the object are m and vrespectively.
Let the mass of the two pieces be m1 and m2respectively after the collision. Let the velocities of the two pieces be v1 andv2 respectively. According to the law of conservation of momentum we have m * v = m1v1 +m2v2 or (m1 + m2) * v =m1v1 + m2v2 where m2 = 3m1 and v1 =0 or (m1 + 3m1) * v = m1 * 0 +3m1v2 or 4m1 * v = 3m1v2 or v2 = (4/3)v The kinetic energy added to the system during theexplosion, with respect to the observer’sreference frame is K = (1/2)m2 * v22 =(1/2)m2 * ((4/3)v)2 =(8/9)m2v2 2i)Let the angle above the horizontal must he throw the ballbe . We know from the relation h = (u2sin2/2g) or sin2 = (2gh/u2) or sin = (2gh/u2)1/2 =((2gh)1/2/u) or = sin-1((2gh)1/2/u) where g = 9.8 m/s2,h = 5.00 m and u = 12 m/s ii)The horizontal distance from the release point to thetarget is R = (u2 * sin(2)/g) iii)The time of flight of the ball is t = (u * sin/g) The speed of the ball just as it reached the target is v = u + gt

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