7. A bullet with a mass of m = 7.50 g is fired with an initial speed of 220 m/s

Question

7. A bullet with a mass of m = 7.50 g is fired with an initial speed of 220 m/s into a block of mass M = 300 g that is initially at rest near the edge of a table of height h = 0.90 m as shown in the figure below. Upon impact, the bullet remains in the block and knocks the block off of the table. Determine the distance d that the block lands away from the base of the table.

8. (BONUS QUESTION) A 1500 kg car travelling west at 25.0 m/s collides head-on with a 9000 kg truck travelling in the same direction but with a speed of 20.0 m/s. The velocity of the truck immediately after the collision is 21.0 m/s to the west. (a) What is the velocity of the car immediately after the collision? (b) How much mechanical energy is lost in the collision?

7. A bullet with a mass of 7.50 g is fired with an initial speed of 220 m/s into a block of mass M-300 g that is initially at rest near the edge of a table of height h 0.90 m as shown in the figure below. Upon impact, the bullet remains in the block and knocks the block off of the table. Determine the distance d that the block lands away from the base of the table. 8. (BONUS QUESTION) A 1500 kg car travelling west at 25.0 m/s collides head-on with a 9000 kg truck travelling in the same direction but with a speed of 20.0 m/s. The velocity of the truck immediately after the collision is 21.0 m/s to the west. (a) What is the velocity of the car immediately after the collision? (b) How much mechanical energy is lost in the collision?

Explanation / Answer

7.

During collision :

vi = velocity of bullet before collision = 220 m/s

m = mass of bullet = 7.50 g = 7.50 x 10-3 kg

M = mass of block = 300 g = 0.3 kg

V = velocity of block -bullet combination after collision

using conservation of momentum

m vi = (m + M) V

(7.50 x 10-3) (220) = ((7.50 x 10-3) + 0.3) V

V = 5.4 m/s

consider the motion along the vertical direction :

Y = vertical height dropped = 0.90 m

a = 9.8 m/s2

t = time = ?

Voy = initial velocity = 0 m/s

using the equation

Y = Voy t + (0.5) a t2

0.90 = 0 t + (0.5) (9.8) t2

t = 0.43 sec

along the horizontal direction :

d = V t = 5.4 x 0.43 = 2.3 m

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