Curling is a sport played with 20 kg stones that slide across an ice surface. Suppose a curling stone sliding at 1.5 m/s strikes another stone and comes to rest in 2.5 ms. Approximately how much force is on the stone during impact. 160N 800N 1600N or 8000N

160N

= 12000 N

Explanation:

Using the formula, we have:

impulse = chenage in momentum

F*t = mV

where,

m = 20kg

t = 2.5ms

= 2.5 * 10⁻³s

v = 1.5m/s

impulse = chenage in momentum

F*t = mV

F = 20*1.5 / (2.5x10^-3)

= 12000 N

The correct answer is not in the options. It is 12000 N

Explanation:

Using one the equations of motion, we have:

v = u + a*t

Where v = final velocity

u = initial velocity

a = acceleration

t = time.

Since the stone comes to rest, v = 0 m/s

=> 0 = 1.5 + (a * 2.5 * 10^ (-3))

=> - 1.5 = 0.0025a

a = - 1.5/0.0025

a = - 600 m/s²

The negative shows that it is a deceleration.

Hence, the force is:

F = m*a

F = 20 * (600)

F = 12000 N