A 2 kg block sits at the edge of a counter that is 1.25 m tall. A 60 gram bullet

Question

A 2 kg block sits at the edge of a counter that is 1.25 m tall. A 60 gram bullet strikes the block becoming embedded in it. The combination shoots off of the counter hitting the floor 1.5 m horizontally from where it started. What was the bullet’s speed just before the collision? A 2 kg block sits at the edge of a counter that is 1.25 m tall. A 60 gram bullet strikes the block becoming embedded in it. The combination shoots off of the counter hitting the floor 1.5 m horizontally from where it started. What was the bullet’s speed just before the collision? A 2 kg block sits at the edge of a counter that is 1.25 m tall. A 60 gram bullet strikes the block becoming embedded in it. The combination shoots off of the counter hitting the floor 1.5 m horizontally from where it started. What was the bullet’s speed just before the collision?

Explanation / Answer

Initial velocity of the bullet block combiation in horizontal direction = Uh

Initial velocity in vertical direction = Uv =zero

Vertical displacement = h = 1.25 m

Time after which ball strikes the ground = t = sq rt (2h / g)

Time after which ball strikes the ground = t = sq rt (2*1.25 /9.8)

Time after which ball strikes the ground = t = 0.5 s

Horizontal distance = R = 1.5 m

Horizontal distance = R = Uh*t

Uh*t =R

Uh=R/t

Uh=1.5 /0.5 = 3 m/s

The bullet block combination was thrown with speed 3 m/s in horizontal direction

Now, momentum conservation

Let V be the velocity of the bullet just before collision

Let m be the mass of the bullet, m = 0.06 kg

and M be the mass of the block, M = 2 kg

mV = (m+M)*Uh

0.06V = 2.06*3

V = 103 m/s

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